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One can show inductively that the Steenrod algebra is generated as an algebra by the $Sq^{2^i}$'s, but can one write down explicitly the Adem relations in terms of this presentation?

Of course, the standard presentation of the Steenrod algebra $\mathcal{A}^\ast$ describes it as the free associative $\mathbb{F}_2$-algebra on the generating set $\{Sq^n \mid n \in \mathbb{N}\}$, modulo the Adem relations plus the identity $Sq^0 = 1$. But it's a standard fact (following from the Adem relations) that this generating set is redundant: $\mathcal{A}^\ast$ is already generated by the restricted generating set $\{Sq^{2^i} \mid i \in \mathbb{N}\}$. However, it's far from obvious how to express a complete set of relations in terms of the restricted generating set; for example, the Adem relations for the product $Sq^{2^i} Sq^{2^j}$ introduce terms of the form $Sq^{2^i+2^j-2^r}Sq^{2^r}$ which must be inductively written in terms of the restricted generating set.

Of course, there are other presentations of the Steenrod algebra, for example in terms of the Milnor basis. But I'm still curious: what are the relations among the $Sq^{2^i}$'s explicitly?

Of course, one could ask a similar question at odd primes.

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This was first addressed by Wall in his 1959 paper: "Generators and Relations for the Steenrod Algebra", C. T. C. Wall, Annals of Mathematics, Vol. 72, No. 3, pp. 429-444. He did not find a simple closed form for the relations, but they look like \begin{align*} \text{Sq}^{2^i} \text{Sq}^{2^j} + \text{Sq}^{2^i} \text{Sq}^{2^j} &= \text{(other terms)} \quad \text{for} \ 0 \leq j \leq i-2, \\ \text{Sq}^{2^i} \text{Sq}^{2^i} &= \text{(other terms)}. \end{align*} Wall is a bit more explicit about the "other terms".

Others have found better descriptions of these. See Section 4.5, and in particular Theorem 4.18, in Wood, "Problems in the Steenrod algebra", Bulletin London Math Soc., Volume 30, Issue 5, pp. 449–517. Wood also mentions a paper by Papastavridis, "Relations in the mod p Steenrod algebra", as dealing with analogues of Wall's results at odd primes, but I haven't read that.

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