# Odd primary dual Steenrod algebra

My question is related to this, this, and this older questions.
Let $\mathcal A_*$ be the dual Steenrod algebra. This is a super-commutative Hopf algebra, and so its $Spec$ is an algebraic super-group.

At  p = 2:

... the algebraic group $Spec(\mathcal A_*)$ can be naturally identified with the group of automorphisms of $Spf(\mathbb F_2[[x]])$ (with $x$ in degree $1$) that preserve:

• the group law $x_1,x_2\mapsto x_1+x_2$,
• the tangent space at the identity.

When p is an odd prime:

... the algebraic super-group $Spec(\mathcal A_*)$ can be naturally identified with the super-group of automorphisms of $Spf(\mathbb F_p[[x,\theta]])$ (with $\theta$ in degree $1$, and $x$ in degree $2$ — this is an exterior algebra on $\theta$ tensor a polynomial algebra on $x$) that preserve:

• the group law $(x_1,\theta_1),(x_2,\theta_2)\mapsto (x_1+x_2+\theta_1\theta_2,\theta_1+\theta_2)$ ????
• the tangent space at the identity. ????

Are these statements correct?
If not, how does one fix them to make them correct?

• When p=2 this is true as stated. When p is odd, maybe try Inoue's paper "Odd primary steenrod algebra, additive formal group laws, and modular invariants" to fill in the question marks? – Dylan Wilson May 2 '18 at 3:12
• It is difficult for me to compare. He works with the formal automorphisms $F$ given by $$F(x)=x+\sum_{n\geqslant1}\xi_nx^{p^n},\qquad F(\theta)=\theta+\sum_{n\geqslant0}\tau_nx^{p^n},$$where $\xi_n$, resp. $\tau_n$ are the polynomial, resp. exterior generators of a Hopf algebra, with diagonal given by composition of such formal automorphisms. I could not find the place where it would be stated that a formal automorphism is of this form if and only if it preserves this or that structure. – მამუკა ჯიბლაძე May 2 '18 at 9:36
• @მამუკა ჯიბლაძე That's very useful. From those formulas, I can see already that the tangent space at the identity is not supposed to be fixed. That tangent space has a two-step filtration, and it's only on the associated graded that the automorphism has to be the identity. – André Henriques May 2 '18 at 10:32
• At one point (a long time ago) I tried quite hard to make this work, and could not come up with a satisfying formulation. But maybe someone else has done better. – Neil Strickland May 2 '18 at 21:39
• There's an exercise appearing in Hopkins's Coctalos notes which asks to show that the odd primary dual steenrod algebra corepresents the strict isomorphisms of the formal additive super group in the category of super algebras. google.co.il/url?sa=t&source=web&rct=j&url=http://… – Saal Hardali Aug 23 '18 at 14:20

This is not a full answer but it was too long for a comment so I decided to write it as detailed answer (i'm sure most if not all of what i'm going to write is already known to you so let me apologize in advance).

I think it helps approaching this story from the other side. Instead of trying to find the correct formal group by staring at the formulas lets try to see if we can squeeze the formal group out from the topology.

Homology is represented by $$H_*(-) = \pi_*(- \otimes H\mathbb{F}_p)$$ from which we see (by virtue of the fact that $$\pi_*$$ is symmetric monoidal over a field) that the hopf algebra $$\pi_* (H\mathbb{F}_p \otimes H\mathbb{F}_p) := \mathcal{A}_*$$ must coact on $$H_*(-)$$. (i will write $$\otimes$$ for smash product of spectra throughout).

So homology determines a symmetric monoidal functor from spectra to graded $$\mathcal{A}_*$$ comodules. And in fact even before taking $$\pi_*$$ we still have (in some vague sense which might be difficult to make precise and I won't need) that $$- \otimes H\mathbb{F}_p$$ lands in the category of comodules over the "derived" hopf algebra $$H\mathbb{F}_p \otimes H\mathbb{F}_p$$.

Interpreting this algebro-geometrically we may think of $$\pi_*(H\mathbb{F}_p \otimes H\mathbb{F}_p)$$ as the ring of functions on some sort of "super" (which just means "in a graded sense") algebraic group $$\mathbb{G} = Spec \mathcal{A}_*$$ and of comodules over it as algebraic representations of $$\mathbb{G}$$. In this language homology determines a symmetric monoidal functor $$\mathcal{Sp} \to Rep(\mathbb{G})$$.

We may now precompose this functor with $$\Sigma^{\infty}_+$$ to get a symmetric monoidal functor (with respect to the product on spaces) $$\mathcal{S} \to Rep(\mathbb{G})$$ (where $$\mathcal{S}$$ stands for the category of spaces). Since every object in $$\mathcal{S}$$ is canonically a cocommutative coalgebra via the diagonal we immediately conclude that the functor actually lands in $$CoAlg(Rep(\mathbb{G}))$$. The algebro-geometric interpretation of coalgebras is as formal schemes by taking $$Spf$$ of the linear dual (there's a caveat here in that the filtration on the dual should come from the filtration on the homology coming from expressing a space as a filtered colimit of finite spaces). Moreover in this form the functor sends products to products since the categorical product in coalgebras is given by the tensor product on the underlying vector space.

Summarizing we now have homology as a functor $$\mathcal{S} \to CoAlg(Rep(\mathbb{G}))$$ which preserves products and filtered colimits. We can also compose with the fogetful $$CoAlg(Rep(\mathbb{G})) \to CoAlg(grVect)$$ which only remembers homology with the coalgebra structure. We may now use this to conjecture about the structure of $$\mathcal{A}_*$$ as follows.

An eilenberg maclane space $$K(A,n)$$ is a commutative group object in $$\mathcal{S}$$. Its homology is therefore a commutative group object in $$CoAlg(Rep(\mathbb{G}))$$. Interpreted algebro-geometrically the homology of $$K(A,n)$$ determines a commutative formal super group $$\mathbb{G}_{A,n}$$ together with a compatible action of the algebraic super group $$\mathbb{G}$$. In other words we get a map which we will hence forth call the comparison map:

$$\mathbb{G} \to Aut(\mathbb{G}_{A,n})$$

Where by $$Aut$$ I mean the full algebraic group of automorphisms of $$\mathbb{G}_{A,n}$$. We may now try to plug in values for $$A$$ and $$n$$ and check whether we get something valuable.

The most obvious first choice is $$A = \mathbb{Z}/p\mathbb{Z}$$ and $$n=1$$. It turns out that this brings us surprisingly close to the correct answer.

Let $$V$$ be an $$\mathbb{F}_p$$ vector space (equivalently a finite $$p$$-torsion abelian group) then there's the following computation:

$$H^{\bullet}(K(V,1),\mathbb{F}_p) = \begin{cases} Sym^{\bullet}(V^{\ast}[-1]) & \text{if p = 2} \\ Sym^{\bullet}(V^{\ast}[-2]) \otimes \bigwedge^{\bullet}(V^{\ast}[-1]) & \text{if p \gt 2} \end{cases}$$

Algebro geometrically this can be interpreted as saying that $$Spf H^{\bullet}(K(V,1),\mathbb{F}_p)$$ is the formal super scheme corresponding to the super vector space $$V \oplus V$$ completed at the origin in the case where $$p$$ is odd and the completion of $$V$$ when $$p$$ is even. It hence inherits the additive formal group structure from the addition on the corresponding vector space. Taking $$V = \mathbb{Z}/p\mathbb{Z}$$ we get something which one might call an additive formal super group.

Now comes the part i'm least sure about.

Its straightforward to compute the automorphism hopf algebra of $$\mathbb{G}_{\mathbb{Z}/p\mathbb{Z},1}$$. Then using the fact that we already know $$\mathcal{A}_*$$ and staring at it for a bit its easy to become convinced $$\mathbb{G}$$ is a sub algebraic-group of $$Aut(\mathbb{G}_{\mathbb{Z}/p\mathbb{Z},1})$$. There are therefore 2 questions left:

1. Find precisely the a subgroup of $$Aut(\mathbb{G}_{\mathbb{Z}/p\mathbb{Z},1})$$ isomorphic to $$\mathbb{G}$$. In other words what kind of extra structure on $$\mathbb{G}_{\mathbb{Z}/p\mathbb{Z},1}$$ do homology co-operation preserve?

2. Analyze the comparison map and determine whether it induces the inclusion from (1)

I think the for the even case we know that (1) is solved by requiring the automorphism be strict and that (2) is an isomorphism (I'm less sure about this point - so i'd love to be corrected about this if i'm wrong). For the case of $$p$$ odd I think the answer to (1) is that one should take automorphisms preserving the even coordinate to first order and the odd coordinate to first order modulo the even coordinate so it preserves the descending degree filtration on the tangent space. I'm less sure about this because I may have made a mistake in the calculation. Regarding (2) it's straightforward that the map must be injective since if a stable cohomology operation in mod p cohqomology is trivial on $$B \mathbb{Z}/p\mathbb{Z}$$ it must be the identity. And finally by computing the action on the cohomology using the axiomatic definition of the steenrod algebra we see that it's the same action.