Maybe people in here have run into this type of structure before:

Is it true that the steenrod algebra at all primes is the rings of function on the automorphism group of the additive formal supergroup at the corresponding prime?

@SaalHardali I think everyone believes it, but you need to make the rhs precise and I never quite saw a satisfying definition of it (what's a "formal supergroup"?)

I think there's a paper of Inoue, let's see if I can find it

I was thinking of this: projecteuclid.org/euclid.jmsj/1149166777, which seems to identify spec of the dual steenrod algebra at odd primes with (something like) Automorphisms of additive fgl's over dual numbers

is there a conceptual interpretation of the gap between this and the fact that the fiber product of the point corresponding to the additive formal group with itself over the moduli stack of formal group is just the automorphisms of the additive formal group?

I mean it seems to indicate naively that the "moduli stack of formal supergroups" could be a better approximation to stable homotopy theory. I gues one should first make sense of a "super complex orientation"...

I'm not sure. I haven't read the paper closely. Of course, there's no gap at p=2 so anything satisfying should account for that

But is there a difference between super and ordinary groups at the prime 2? I'm pretty sure that the super linear algebra is just ordinary linear algebra at 2

Maybe that's an old misconception of mine though...

(I haven't read the paper either yet)

but (as you may know) any satisfying algebrogeometric interpretation of the odd primary steenrod algebra (or its dual, rather) should account for the odd degree stuff (the exterior factors). I'm pretty sure it is the case, and maybe Inoue says as much, that Aut of the additive group is spec of the even part, like at p=2

I see.

@DenisNardin I'm guessing your question on my post means you're thinking about Galois extensions?

Well, in your situation B is a bundle of G-representations on A (let's suppose H=O(G) for some algebraic group G), I'm just trying to see if we can say things about the fibers

@DenisNardin Yeah I guess I'm not sure. What is the torsor condition supposed to mean here?

Does that tell me that the fiber looks like G?

Hey @Aly!

I mean, it is a representation where $V\otimes V \cong V\otimes \rho$ where $\rho$ is the regular representation

Oh I just realized you may not be able to talk in here Aly, because of MO's rules about how much rep you have to have.

If you interpret it via characters (let's pretend we're in char 0 ) this is saying $\chi^2=n\chi$, so I assume this means $V$ must be the regular representation?

@DenisNardin I'm not sure. I imagine this is an elementary representation theory problem though that I don't know how to do off the top of my head.

Oh... yeah ok I guess I can piece that together.

@Aly I tried to give you explicit write-access to this room, let me know if that worked.

Hey, thanks @JonathanBeardsley. Just lurking :)

@Aly Ah ok, sorry to draw attention! Lurk away. =P

@SaalHardali i feel extremely strongly that formal supergeometry does not have much to say here. every place where i have found someone suggesting that some odd classes can be explained by formal supergeometry, i have been able to find a more satisfying explanation for what's going on in terms of a resolution

for example: you can compute the cohomology of BC_p using the serre spectral sequence / gysin sequence associated to the spherical fibration S^1 –> BC_p –> CP^∞, and what you end up learning is that there's a long exact sequence with nodes E^* CP^∞ –(- cup [p](x))-> E^* CP^∞ –> E^* BC_p, where "[p]" is induced by the multiplication by p map on CP^∞ = K(Z, 2) and x is your choice of orientation class

for a huge family of cohomology theories, including all morava K-theories, this class is not a zero-divisor, and you end up learning that E^* BC_p = E^* CP^∞ / [p](x) as a quotient of rings, and this has algebro-geometric meaning by way of the p-torsion of the formal group Spf E^* CP^∞

in the case of E = HFp, this class [p](x) = px = 0 is accidentally zero, and this same long exact sequence forces the existence of an odd class into HFp^* BC_p, recording the fact that you did a silly thing and tried to quotient by zero

all the odd-degree classes in the steenrod algebra can be viewed as artifacts arising from beginning with HFp_* BP and quotienting, bit by bit, down to HFp_* HFp, where the map you're coning off acts by zero on homology

i don't want to claim that this is universally applicable—i don't know how to explain eta this way, say, or that i would want to—but in all formal-geometry-adjacent situations this seems to be (perhaps tautologically) true

@EricPeterson Thanks, i'm not sure I'm qualified/experienced enough to understand you answer though. BTW your book (or at least the begining) is very good! Thanks for that too!

all i mean to say is that the reason bocksteins are called bocksteins in topology is reflected also in the algebraic geometry. no need to find a new framework

sorry if i over-answered; this was a real source of confusion in grad school for me—and eventual pet peeve when i felt i had it figured out. other bat signals i'll come running for to naysay include "does the E_infty ring MU classify derived formal groups?"

@EricPeterson only the E_2-ring

your answer was helpful to me, even if i wasn't intended

(& thank you for the book compliment :) )

@PeterNelson That's a nice paper, essentially it describes the group (super)scheme represented by the dual Steenrod algebra. In a sense it's only half the work though: we'd like to see that group as the automorphism group of a geometric object. Anyway, I'll heed the wise words of Eric (even if there must be such an object, dammit!)