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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 $p \neq 2$?

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The cohomology $H^*(\mathbb{RP}^\infty)$ carries a formal group law, determined (as in the complex case) by the rule for taking the tensor product of two real line bundles. It's the additive formal group law. Co-operations on cohomology determine, therefore, natural operations on this formal group law. When $p$ is odd my recollection is that there is a graded-commutative version but I do not remember enough to say things with confidence. – Tyler Lawson Dec 10 '11 at 5:01
My recollection is that there were some course notes out there online that explained this pretty well, but my memory is not functioning well here either. Maybe someone else knows a link? – Tyler Lawson Dec 10 '11 at 5:02
@Tyler, perhaps you're thinking of Pages 22-23 of Hopkins COCTALOS notes: – Justin Noel Dec 10 '11 at 8:12
Note also Geoffrey Powell, "Unstable modules over the Steenrod algebra revisited", – Charles Rezk Dec 10 '11 at 15:28
@Tyler: I learned this from Lurie's notes on chromatic homotopy theory, – Akhil Mathew Dec 10 '11 at 16:24
up vote 13 down vote accepted

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.

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Thanks a lot. This was a philosophical question, and I enjoyed reading this philosophical answer. – Akhil Mathew Dec 12 '11 at 3:14
Incidentally, how do the computations you list show that the stable homotopy category cannot be a derived category of some dga? – Akhil Mathew Dec 12 '11 at 3:21
@Akhil: For one, there is no way to put a ring structure on the "mod-2 sphere". For any DGA $R$ you can derived-tensor with $\mathbb{Z}/2$ and get a an exact triangle $R \to R \to R/2$ where $R/2$ is an associative DGA where the multiplication-by-2 map is zero. This can't happen in stable homotopy theory; the mod-2 sphere doesn't admit a multiplication. – Tyler Lawson Dec 12 '11 at 5:28

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.

-- edit --

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.

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I don't think there is something that primal yet available; instead, we have this deep sea of computations that permit many guesses, some of which turn out to be true. For instance, MU is designed to be universal among complex-oriented spectra with a choice of coordinate on the formal group. You could hope that the constraint "among complex-oriented spectra" is empty, and instead MU carries a coord. universal among all formal groups. That this is true is another shocking freeness result, but we don't yet have a construction of a spectrum that starts with that data from scratch, so to speak... – Eric Peterson Dec 10 '11 at 20:31
@David: Thanks for doing that! I've been trying to get somewhere in Lurie's notes, but I think his proof is essentially a slightly fancier restatement of the usual proof (i.e. the Adams spectral sequence computation, except that he only carries it out at $p = 2$). In particular, one of the things that feels somehow "morally wrong" about it (to me, at least) is that the result seems to fall out by coincidence: that is, $\pi_* MU$ is computed to be a polynomial ring that happens to match the Lazard ring, and the map $L \to \pi_* MU$ is checked on indecomposables to match. – Akhil Mathew Dec 10 '11 at 23:42
(contd.) Yet $MU$ is universal among complex-oriented cohomology theories, and the fact that its formal group law is the universal one seems more than mere coincidence. But of course I'm a beginner to this field and my opinions should be taken very seriously... – Akhil Mathew Dec 10 '11 at 23:44
@Akhil: Yes that's the one. The one with the more obvious title seems to just be an announcement of results. On another note, it sounds like you're expressing a purely formal groupy way of constructing MU. I don't think such a thing is written down but I'm tempted to think one could attempt the same idea as in Lurie's work on elliptic cohomology: define what should be a "formal group" or "derived formal group" over an E_infty ring, then define the obvious functor, show that it's represented by something reasonable, and take global sections. The only part that I think would be awkward is that – Dylan Wilson Dec 10 '11 at 23:54
(contd) the thing doing the representing would have to be like an infinite-dimensional artin stack... and I actually haven't seen the exact hypotheses on Lurie's representability theorem, so I don't know if you could apply it in this case. He says he's trying to post this before the new year, but until then the only reference I know is his original thesis, but this treats the case where E_infty rings are replaced by SCRs, so I'm not sure if the same proof applies. There also finite-dimensionality hypotheses... – Dylan Wilson Dec 10 '11 at 23:55

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