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## Computing squaring operations in the Adams spectral sequence

This question is about the classical Adams spectral sequence. Squaring operations are defined on its $E_2$ term. I'd like to know how to compute some of the non-trivial operations, such as $Sq^2 ( c_0 ) = h_0 e_0$. I feel like this ought to be doable in the May spectral sequence, but I don't know the details.

I'm aware of some work of Milgram on the subject, but there are some problems with his approach because of indeterminacies of Massey products.

Thanks!

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Congratulations on asking the 1000th algebraic topology question! – Dan Ramras Mar 23 2011 at 16:54
I think that Chris Douglas and Mike Hill may have thought about something along these lines. I recall a talk that Chris gave about it. I don't think it has been written up. The title of the talk was something like "Juggling Toda Brackets." I suspect what they were doing was at least related to Milgram's work (they were finding ways to deal with indeterminacies in Toda brackets/Massey products). I probably have notes from the talk at home, so if Mike and Chris have forgotten about it, send me an e-mail and I can dig up my notes. – Dan Ramras Mar 23 2011 at 16:58
Bob Bruner knows an enormous amount about this, and has computer programs that can do all sorts of useful things. I think that most of this is unpublished. You should ask him. – Neil Strickland Mar 23 2011 at 18:29
Because of Bob Bruner's computer calculations, I know what the answers ought to be. But I'm looking for a conceptual way of obtaining the answers. – Dan Isaksen Mar 25 2011 at 13:51
I think what Dan would like is a precise statement about the relation between the Sq^i in Ext_A and the Sq^i in Ext_{E^0(A)}, the E_2 term of the May ss. The problem with brackets is indeterminacy. Of course, May ss, or any ss, calculations will suffer the same problem. – Robert Bruner Mar 27 2011 at 3:03

I think for $k>0$ currently no-one knows an efficient algorithmic way to compute the $Sq^k$, e.g., from a minimal resolution. (The $Sq^0$ is easy since it is induced by the "Frobenius" map on $A_\ast$ and one just needs to compute a chain map).

For the May spectral sequence you might want to check out Nakamura, Osamu. On the squaring operations in the May spectral sequence. Mem. Fac. Sci. Kyushu Univ. Ser. A 26 (1972), no. 2, 293–308 (Google can find this online).

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 There is one efficient 'algorithm' for the Sq^i other than Sq^0, and it is due to Nassau! It does require an actual resolution, not just a spectral sequence, and Dan was hoping to avoid this. I will post a brief account as a separate answer because I don't have much room here and can't make it display math here except as raw TeX code. – Robert Bruner Mar 26 2011 at 20:24 I have been trying to make my computer brute force these for months, but it has been unbearably slow. This is bad news for me, but explains why I have been thus far unable to do these computations quickly. – Joseph Victor Nov 2 at 18:40

You can avoid (some) indeterminacies by using Baues's secondary Steenrod algebra and it's relation to the $E_2$ and $E_3$ terms of the Adams spectral sequence:

MR2220189 (2008a:55015) Baues, Hans-Joachim The algebra of secondary cohomology operations. Progress in Mathematics, 247. Birkhäuser Verlag, Basel, 2006. xxxii+483 pp. ISBN: 3-7643-7448-9; 978-3-7643-7448-8 (Reviewer: David Chataur), 55S20 (18G10 55T15)

MR2193337 (2006k:55031) Baues, Hans-Joachim; Jibladze, Mamuka Secondary derived functors and the Adams spectral sequence. Topology 45 (2006), no. 2, 295–324. (Reviewer: J. P. C. Greenlees), 55T15 (18D05 18G10 55S20 55U35)

So far, this has been successfully applied to the computation of tons of Massey products (see the first reference), but one can also deal with squaring operations along the lines of this paper and

MR2482810 (2009i:55015) Baues, Hans-Joachim; Muro, Fernando Toda brackets and cup-one squares for ring spectra. Comm. Algebra 37 (2009), no. 1, 56–82. (Reviewer: Tyler D. Lawson), 55Q35 (55P43)

Sorry for this shameless self-propaganda, and also for not being more explicit here, but I find no way to answer this question in a short enough but explicit way.

P.D. I know this should have been a comment, but it is too long for that.

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 Fernando, What would be needed to get cup-2 and higher by these methods? – Robert Bruner Mar 27 2011 at 10:13

Suppose $x \in Ext^{s,t}_A(k,k)$ and let $\mathcal{C} = \cdots \to C_s \to \cdots \to C_0 \to \to k \to 0$ be a resolution so that $x$ is represented by a cocycle $C_s \to \Sigma^t k$.

To compute $Sq^i(x)$, find a 'small' extension $0 \to \Sigma^t k \to M_{s-1} \to \cdots \to M_0 \to k \to 0$ realizing x and call it $\mathcal{X}$. Then $x$ is represented by a chain map $\mathcal{C} \to \mathcal{X}$ with the cocycle $C_s \to \Sigma^t k$ at one end and the identity of $k$ at the other. The cocommutative Hopf algebra structure of $A$ makes $\mathcal{X} \otimes_k \mathcal{X}$ a complex of $A$-modules, and there is a chain map $\chi : \mathcal{C} \to \mathcal{X} \otimes_k \mathcal{X}$ covering $1_k$ whose other end $C_{2s} \to \Sigma^{2t} k$ is a cocycle representing $x^2$. The composite $\tau \chi$ of this chain map with the twist map $\tau : \mathcal{X} \otimes \mathcal{X} \to \mathcal{X} \otimes \mathcal{X}$ is another lift of $1_k$, so there is a chain homotopy $\chi_1 : \mathcal{C} \to \mathcal{X} \otimes \mathcal{X}$ between them. This gives a cocycle $C_{2s-1} \to \Sigma^{2t} k$ which represents $x \cup_1 x = Sq^{s-1}(x)$. Repeat to get $\chi_2$, giving a cocycle $C_{2s-2} \to \Sigma^{2t} k$ representing $x \cup_2 x = Sq^{s-2}(x)$, etc. If the extension $\mathcal{X}$ is small, this is a remarkably effective way to compute these operations, ${\it{\text{ if you have a resolution to work with}}}$.

Two minute exercise: show that $Sq^0(h_0) = h_1$ in $Ext_{A(1)}(F_2,F_2)$ using this method and the resolution

$C_1 = \Sigma^1 A(1) \oplus \Sigma^2 A(1) \to C_0 = A(1)$ by $\iota_1 \mapsto Sq^1$, $\iota_2 \mapsto Sq^2$, and

$C_2 = \Sigma^2 A(1) \oplus \Sigma^4 A(1) \to C_1$ by $\iota_2 \mapsto Sq^1 \iota_1$ and $\iota_4 \mapsto Sq^1 Sq^2 \iota_1 + Sq^2 \iota_2$.

What have we done? The universal case is a $C_2$-equivariant map $\mathcal{W} \otimes \mathcal{C} \to \mathcal{C} \otimes \mathcal{C}$ extending the diagonal $\mathcal{C} \to \mathcal{C} \otimes \mathcal{C}$ coming from the cocommutative coproduct of $A$. This is what May's Springer LNM V. 168 article needs to construct Steenrod operations. Unfortunately, if $\mathcal{C}$ is a production strength resolution, not just some little toy, then $\mathcal{C} \otimes \mathcal{C}$ is a monster, and we do not want to be trying to lift maps over its differential. (Maybe you, dear reader, see a way out of this. If so, this is great news!)

Christian's clever observation is that we can do this one cocycle at a time, and compute the map $\mathcal{W} \otimes \mathcal{C} \to \mathcal{X} \otimes \mathcal{X}$ if we can find small extensions $\mathcal{X}$ representing the cocycles of interest. This step is not algorithmic, because the easy answers are too large, essentially as large as $\mathcal{C}$ itself. On the other hand it is quite feasible by hand in low degrees. Sean Tilson and I have carried this out for $c_0$, $d_0$, $e_0$, $f_0$ and $n$ (in the BMT notation for $Ext_A(F_2,F_2)$), perhaps another one or two. I suppose we should convert the results from computer code to TeX one day. I was interested in this in order to unambiguously determine which of the two possible elements in $Ext^{4,22}$ found by my computer program was $f_0 = Sq^1(c_0)$ and which was $f_0 + h_1^3 h_4$. Similarly for $f_1 \in Ext^{4,44}$; this follows by $Sq^0$ from identifying $f_0$, and as Christian notes, $Sq^0$ is easy since that is induced by the Frobenius in the dual of $A$, hence by restriction along the dual of the Frobenius, $A \to D(A)$, where $D(A)$ is the 'double' of $A$, $D(A)_{2n} = A_n$ and $D(A)_{2n+1} = 0$. On Milnor basis elements this is $Sq(2r_1, \ldots, 2r_k) \mapsto Sq(r_1, \ldots, r_k)$ while basis elements with any odd entries go to 0, so it is easy to automate.

Finally, since what I thought was going to be a paragraph or two has already grown absurdly long, I might as well go whole hog and mention a favorite problem. The complex $\mathcal{X} \otimes \mathcal{X}$ is one way to construct a complex representing $x^2$. The Yoneda composite of $\mathcal{X}$ with itself is a much 'slimmer' one. Unfortunately, we cannot recognize $\tau$ in it. If we could, we could then automate the calculation of the $Sq^i$. This is the problem of a description of the squaring operations in terms of Yoneda product rather than tensor product and a solution would be great to have.

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This is an answer Bob's question after my previous answer. It doesn't fit in a comment.

As a spectrum, $\mathrm{End}(H\mathbb{Z}/2)$ is known to be a product of Eilenberg-MacLane spectra, so it is the classifying spectrum $\mathbb{H} C_{*}$ of a chain complex $C_{*}$, i.e. it is in the image of the Eilenbeg-MacLane functor $\mathbb{H}$ from chain complexes to spectra. The work of Shipley shows that this functor is well behaved with respect to multipliciative structures, since the symmetric monoidal model category of chain complexes is Quillen equivalent to that of $H\mathbb{Z}$-algebra spectra. In particular $\mathbb{H}$ takes chain algebras to ring spectra. However, we know that the ring spectrum $\mathrm{End}(H\mathbb{Z}/2)$ is not an $H\mathbb{Z}$-algebra, so it is the classifying spectrum of a chain complex, but not the classifying ring spectrum of a chain algebra.

Your question would be answered in full generality if we understood what a ring spectrum structure over the classifying spectrum of a chain complex is, i.e. what structure on a chain complex $C_{*}$ is equivalent to an $S$-algebra structure on $\mathbb{H} C_{*}$.

This is an achievable task, since there are nice models for the functor $\mathbb{H}$ taking values on symmetric spectra, and if $X$ is a symmetric spectrum, it is easy to describe a ring spectrum structure in terms of maps of simplicial sets $X_p\wedge X_q\rightarrow X_{p+q}$. The outcome should be some sort of non-additive multiplication on the chain complex $C_*$.

Now, let me remind you that Leif Kristensen constructed a long time ago a chain complex $C_{*}$ whose cohomology is the Steenrod algebra (reference below). This chain complex is endowed with a multiplication, which induces the product on the Steenrod algebra in cohomology. However, Kristensen's multiplication is not a chain algebra structure on $C_{*}$ since it is left distributive but not right distributive. This could be an approximation to a solution to the previous problem.

So far, I've been very lazy to pursue this project. There'd be a lot of work to do, but the outcome can be very valuable and surprising, and it looks like a feasible project.

MR0159333 (28 #2550) Kristensen, Leif On secondary cohomology operations. Math. Scand. 12 1963 57–82. (Reviewer: J. F. Adams)

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 Fernando, This is great. I'm sorry I only now saw it. I guess I didn't get notified because it was a new answer rather than a comment. Anyway, Thanks! There are other interesting operations that are distributive on one side only, with some kind of additional structure on the other side: unstable composition and H_infinity homotopy operations are the ones I know. Bob – Robert Bruner Feb 28 2012 at 0:59