# Why does the Steenrod algebra act faithfully on $H^\ast(BC_p)$?

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.

• 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 $x^{2^k}$. Dec 12 '19 at 17:33
• @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... Dec 12 '19 at 17:40
• 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. Dec 12 '19 at 17:49
• 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 $sq^1 (x^2)=0$. Dec 12 '19 at 22:39
• @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. Dec 12 '19 at 22:40

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.
• @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 $\Bbb Z/p$. Dec 13 '19 at 20:59