Is there a high-concept explanation of the dual Steenrod algebra as the automorphism group scheme of the formal additive group?

Question

Recall that for any space $X$, the cohomology $H^*X$ (always, in this post, with $\mathbb{Z}/2$-coefficients) has an action of the Steenrod algebra $\mathcal{A}$; that is, a natural morphism $\mathcal{A} \otimes H^*X \to H^*X$. This is not a morphism of algebras, but $\mathcal{A}$ has a Hopf algebra structure such that the appropriate diagrams involving comultilication on $\mathcal{A}$ and $H^*X$ commute.

Consider now a space with finitely generated homology in each degree. Then the Steenrod algebra acts on homology by duality, and dualizing this gives that the dual Steenrod algebra $\mathcal{A}^{\vee}$ coacts on the completed cohomology ring; that is, there is a map $$ H^*X \to H^*X \hat{\otimes} \mathcal{A}^{\vee}$$ which is in fact a morphism of rings, and makes appropriate diagrams commute for products in $X$. It follows that when $X$ is a H space, and $H^*X$ is a Hopf algebra, then we have a coaction of $\mathcal{A}^{\vee}$ on the completed cohomology ring $H^{\star \star}X$ (or something like that). Anyway, the upshot of this is that $\mathrm{Spec} \mathcal{A}^{\vee}$ is a (noncommutative) group scheme because of the Hopf algebra structure, and what we really have is an action of this group scheme (in some sense, at least) on the formal scheme $\mathrm{Sppf} H^{\star \star}(X)$.

In the case where $X = \mathbb{RP}^\infty$, then the formal scheme just described is the formal additive group. Milnor's paper "The Steenrod algebra and its dual" shows that $\mathrm{Spec} \mathcal{A}^{\vee}$ is precisely the automorphism group scheme of this (i.e. a polynomial ring on variables in each power of $2$ minus one). This is established by computation in Milnor's paper.

Q1: Is there a high-concept explanation of why this should the case?

Q2: What's the analog in characteristic $p \neq 2$?

Answer 1

What I'd like to say is, to some degree, commentary and expansion on what some people have said in the comments above.

I feel that the answer to (Q1), about whether there is a "high-concept" explanation for the dual Steenrod algebra, is "no" at the current point in time. Even further, I feel like the attempt to do so right now would be misleading. We can assign some high-concept descriptions to phenomena in terms of formal group laws, but trying to give a high-concept reason why the Steenrod algebra takes the form it does would be misinterpreting the role of the Steenrod algebra.

We construct the stable homotopy category from the category of topological spaces. Based on its construction, we can say a lot of things in older and more modern language. It's a tensor triangulated category; it's the homotopy category of a symmetric monoidal stable model category; it's the homotopy category of a stable $\infty$-category; it's some kind of "universal" way to construct a stable category out of the category of topological spaces. Based on this, we can describe a lot of the foundational properties of the stable homotopy category and the category of spectra. However, there are a ton of categories that satisfy basically the same properties. Without doing more work, we don't understand anything about specifics that distinguish the stable homotopy category from any other example.

These kinds of computations originate with the Hurewicz theorem, the Freudenthal suspension theorem, Serre's computation of the cohomology of Eilenberg-Mac Lane spaces and his method for computing homotopy groups iteratively, and Adams' method of stepping away from Postnikov towers and upgrading Serre's method into the Adams spectral sequence. Before these, you would have no reason to necessarily believe that the stable homotopy category isn't equivalent to, say, the derived category of chain complexes over $\mathbb{Z}$ or some other weird differential graded algebra. Before you have these, you don't have Milnor's computation of the homotopy groups of $MU$ or $MU \wedge MU$, and you don't have Quillen's interpretation in terms of the Lazard ring.

In short, understanding the Steenrod algebra and its dual are prerequisites for all the qualitative things we understand about the stable homotopy category to distinguish it from another example. Right now there is not a door to the stable homotopy category that comes from formal-group data, as much as we would like one. As a result, I guess I feel like trying to assign a high-concept description to the Steenrod algebra is backwards right now.

The fact that often computations come first and the conceptual interpretations second is something that gives the subject some of its flavor. Are there high-concept explanations why vector bundles should have Stiefel-Whitney classes? Why complex vector bundles should have Chern classes? Why the "quadratic" power operations in mod-$2$ cohomology should generate all cohomology operations, and why all the relations are determined by those coming from composing two?

This is definitely not to say that such a description wouldn't be desirable. It would be very desirable to have a more direct route from concepts to the stable homotopy category, because constructing objects that realize conceptual descriptions can be very difficult.

Answer 2

Given two spectra $E$ and $F$, how might we get a handle on the contents of the spectrum $E \wedge F$? One thing we could try is to produce an interesting map involving $E \wedge F$ as its source or target that relates to things we already understand. Let's assume complex orientations for $E$ and $F$, so maps $u: MU \to E$ and $v: MU \to F$, which we combine to get a map $MU \wedge MU \to E \wedge F$. The homotopy of the spectrum $MU \wedge MU$ carries the universal example of a formal group law isomorphism, and so the map we constructed selects an isomorphism of the formal group laws associated to the composite orientations $MU \xrightarrow{u} E \xrightarrow{\eta_F} E \wedge F$ and $MU \xrightarrow{v} F \xrightarrow{\eta_E} E \wedge F$. In some cases we are lucky enough to produce an isomorphism of the formal groups of $E$ and of $F$, like with your computation of interest when $E = F = H\mathbb{F}_2$: $$\operatorname{Spec} \pi_* H\mathbb{F}_2 \wedge H\mathbb{F}_2 = \operatorname{Aut}(\hat{\mathbb{G}}_a).$$ This is definitely not going to happen in general, since the homotopy of the smash product can be either too complicated or not complicated enough, or the composite orientations might not compare well with the originals --- they might be "damaged" in some way through pushforward. In the case of the dual Steenrod algebra it is in some sense a statement about spread-out-ness of the objects involved. But, as far as I know, this is as deep an explanation as you can get at present.

I'm told that something similar happens in the odd primary case, but involves automorphisms of the formal additive "supergroup". I have no idea how it works, though, and is probably generally related to my poor understanding of graded commutativity and odd dimensional phenomena in general. Definitely it is mentioned in brief in COCTALOS; searching on 'super' will bring it right up if you want to read a few sentences more.

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Your mentioning of $\mathbb{R}\mathrm{P}^\infty$ is somewhat separate. The reason the Steenrod operations show up there is that $\mathbb{R}\mathrm{P}^\infty$ is a $B\mathbb{Z}/2$ and $\mathbb{Z}/2 = \Sigma_2$ is a symmetric group permuting the inputs to a cup product. It's an altogether different miracle that this recipe for constructing cohomology operations exhausts all of $H\mathbb{F}_2$'s.