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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 presentation of the mod 2 Steenrod algebra $A$ with generators $s_i$ corresponding to $\mathrm{Sq}^{2^i}$ and relations of the form $$ [s_i,s_j] = T_{ij}\quad 0\leqslant j\leqslant i-2$$ $$s_i^2 +s_is_{i-1}s_i+s_{i-1}^2s_i = T_i $$ where $T_{ij}$ and $T_i$ belong to the subalgebra generated by $s_0,\ldots,s_{i-1}$. Using this presentation Wall was able to solve a conjectured posed earlier by Toda.

Question 1 Has anyone worked out the terms $T_{ij}$ and $T_i$ for low values of $i$ and $j$? I presume that the terms in $T_{ij}$ and $T_i$ are at least cubic in the $s_i$? Is this true? (I'll re-read the paper with a bit more care and probably figure this out, but perhaps someone can confirm this quickly).

Of course this gives presentations of the subalgebras $A(n)$ generated by $s_0,\ldots,s_n$ in $A$. It is easy to see, for example, that

  • $A(0) = T(x\mid x^2)$
  • $A(1) = T(x,y\mid \text{add } y^2 + xyx)$.
  • $A(2) = T(x,y,z \mid \text{add } [x,z]+yxy, z^2+y[z,y])$

In the lovely User's guide to spectral sequence J. McCleary uses the first two presentations above to do some computations in the Adams spectral sequence: it turns out $A(0)$ admits a $(4,12)$ periodic resolution as an $A(1)$-module. A particular case of my question is:

Question 2 Has anyone worked out a presentation of say, $A(3)$? Is it known if each $A(n)$ is finite dimensional? In this case, does one know how quickly $\dim A(n)$ grows like, for example?

I am definitely not well-versed in the story of this algebra, so if the question is answered by either a paper, or book, or if there is literature that points in the right direction, I am happy to keep reading that in lieu of a full answer.

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    $\begingroup$ Disclaimer: I presume that with some spare time I can compute the presentation of $A(2)$ suggested by Wall. My question is not "can this be done?" but rather "has this been done systematically enough for low values of $n$?" $\endgroup$
    – Pedro
    Commented Jun 22, 2020 at 0:46
  • $\begingroup$ Add: I added the presentation of $A(2)$. $\endgroup$
    – Pedro
    Commented Jun 22, 2020 at 1:51

1 Answer 1

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Question 1: "Closed forms" for the relations are known, due to Grant Walker in what seems to be unpublished work. The relations are described in Wood's paper "Problems in the Steenrod algebra," (PDF) Theorem 4.18. For example, $$ s_i^2 = s_{i-1} \chi(s_i) s_{i-1} + s_{i-1}^2 \chi(s_i), $$ where $\chi$ is the antipode. I put "closed" in quotes because this reduces the problem to having a closed form for $\chi(s_i)$. Wood gives some information about that, too, and in any case, it's easy to compute for small values of $i$ either by hand or using SageMath.

Wood's paper is a nice place to find lots of information about the Steenrod algebra; I recommend it.

Question 2: Certainly each $A(n)$ is finite-dimensional: by Milnor's famous theorem about the dual of the Steenrod algebra, $A(n)$ is dual to a quotient of a polynomial ring, namely $$ \mathbb{F}_2 [\xi_1, \xi_2, \xi_3, ...] / (\xi_1^{2^{n+1}}, \xi_2^{2^n}, ..., \xi_i^{2^{n+2-i}}, ..., \xi_{n+1}^2, \xi_n, \xi_{n+1}, ...). $$ So its vector-space dimension is $2^\binom{n+2}{2}$.

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  • $\begingroup$ Dear John: thank you, this is really useful. I think I almost opened that paper a few days ago! $\endgroup$
    – Pedro
    Commented Jun 22, 2020 at 2:12
  • $\begingroup$ I was just trying to access this link idealibrary.com/links/citation/0021%2d8693/188/203 in your website but for some weird reason it redirects to oladieta.com.br. I thought you should know. $\endgroup$
    – Pedro
    Commented Jun 24, 2020 at 13:54

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