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Define the Steenrod algebra $A^\ast$ to be the algebra of all stable mod $p$ cohomology operations. Without actually computing $A^\ast$, is it possible to see that $A^\ast$ acts faithfully on $H^\ast(BC_p; \mathbb F_p)$?

My question is closely related to this one.

Here is an attempt at an argument. (EDIT: As Tyler's comment shows, this argument doesn't work! I'll leave it up, though, as an example of the kind of thing I might hope could be true) Let $\phi: H \to \Sigma^r H$ be a nonzero stable cohomology operation (where $H = H\mathbb F_p$ is the Eilenberg-MacLane spectrum). Then

$$\phi \wedge 1 : H \wedge H \to \Sigma^r H \wedge H$$

is a nonzero $H$-module map, and so is nonzero on homotopy. We have $H = \varinjlim_n \Sigma^{-n} K(\mathbb F_p, n)$ . It follows that

$$\phi \wedge 1: H \wedge \Sigma^{-n} K(\mathbb F_p, n) \to \Sigma^r H \wedge \Sigma^{-n} K(\mathbb F_p, n)$$

is nonzero on homotopy for some $n$. I think it's the case that $H_\ast(K(\mathbb F_p, n))$ is generated under Pontryagin product by $H_\ast(\Sigma K(\mathbb F_p, n-1))$ -- but I'm not sure if this is true, much less whether there is a non-computational reason for it. This ought to allow us to induct downwards to show that

$$\phi \wedge 1: H \wedge K(\mathbb F_p, 1)^N \to \Sigma^r H \wedge K(\mathbb F_p, 1)^N$$

is nonzero on homotopy for some $N$, which is almost the desired conclusion.

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    $\begingroup$ Unfortunately it's not the case that $H_* (K(\Bbb F_p, n))$ is generated under the Pontrjagin product by suspended classes from $H_*(K(\Bbb F_p, n-1))$. For example, $H^*(K(\Bbb F_2, 2))$ is a polynomial algebra on elements $x, Sq^1(x), Sq^2 Sq^1(x), \dots$ that are all primitive under the coproduct. When you take duals, you get a divided power algebra, which is an exterior algebra on classes dual to $x^{2^k}, (Sq^1(x))^{2^k}, (Sq^2 Sq^1(x))^{2^k}, \dots$ -- the suspended classes only cover the $x^{2^k}$. $\endgroup$
    – Tyler Lawson
    Commented Dec 12, 2019 at 17:33
  • $\begingroup$ @TylerLawson Thanks! But I thought that Thm 8.11 of Wilson's BP sampler was saying that $H_\ast(K(F_p, n))$ is the tensor product of an exterior algebra and a truncated polynomial algebra, rather than a divided power algebra? Admittedly, I'm very unsure of how to read results involving Hopf rings... $\endgroup$
    – Tim Campion
    Commented Dec 12, 2019 at 17:40
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    $\begingroup$ You're correct. But in characteristic p, a divided power algebra on x turns into a tensor of truncated polynomial algebras generated by x^{p^k}/(p^k)!, because if y is an element, then y^p = p! (y^p/p!) == 0. And thanks to the magic of characteristic two, these truncated polynomial algebras are also exterior algebras. $\endgroup$
    – Tyler Lawson
    Commented Dec 12, 2019 at 17:49
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    $\begingroup$ Tim, could you clarify for me what you mean by acting faithfully? For example, how can $\mathscr{A}$ be said to act faithfully on $H^*(K(\mathbb{F}_2,1),\mathbb{F}_2)=\mathbb{F}_2 [x]$ when $sq^1 (x^2)=0$. $\endgroup$
    – Connor Malin
    Commented Dec 12, 2019 at 22:39
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    $\begingroup$ @ConnorMalin I mean that for every nonzero $\phi \in A^\ast$, there exists $\alpha \in H^\ast(BC_p)$ such that $\phi(\alpha) \neq 0$. So the map $A^\ast \to Hom(H^\ast(BC_p),H^\ast(BC_p))$ is injective. $\endgroup$
    – Tim Campion
    Commented Dec 12, 2019 at 22:40

1 Answer 1

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As is commented by @Connor Malin, the action of the Steenrod algebra on $H^*B\mathbb{Z}/p$ is not faithful. Consider the case $p=2$. $Sq^3Sq^1$ acts trivially on $H^*(B \mathbb{Z}/2)$, since $Sq^{2n+1}x^{2m}=0$ by the Cartan formula. As a matter of fact, the computation of the Hopf ring structure of $H_*K(\mathbb{Z}/p,*)$ shows that for no finite $n$, the action of the Steenrod algebra on $H^*((B\mathbb{Z}/p)^n)$ can be faithful. One can see this more easily using the last section of T.Kashiwabara Hopf Rings and Unstable Operations, JPAA 94 (1994) 183-193, or even from classical computations of Steenrod algebra.

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  • $\begingroup$ Thanks! Now I’m very confused because in some sense the coaction of the dual Steenrod algebra is “faithful”— it’s used to construct the isomorphism between spec of the dual Steenrod algebra and a certain automorphism scheme of the additive formal group... $\endgroup$
    – Tim Campion
    Commented Dec 13, 2019 at 14:25
  • $\begingroup$ As a matter of fact the action is "faithful on generaors", which somewhat dualizes to what you are saying. $\endgroup$
    – user43326
    Commented Dec 13, 2019 at 14:39
  • $\begingroup$ @TimCampion I think that the difference is that your question is about the literal action of $A^*$ on the ring $H^*(B\Bbb Z/p)$, whereas the coaction is scheme-theoretic--it describes an action of $A^*$ on the $R$-points of the additive formal group for any extension ring $R$ of $\Bbb Z/p$. $\endgroup$
    – Tyler Lawson
    Commented Dec 13, 2019 at 20:59

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