Suppose $C$ is a nice stable $\infty$-category, let $a\Delta^{op}$ be the augmented simplex category. There's a functor $M: Fun(a \Delta^{op}, C) \rightarrow \prod Fun(\Delta^{1}, C)$ from augmented simplicial objects into the product of arrow $\infty$-categories, sending $(X_{\bullet})$ to the "matching maps" $X_{0} \rightarrow X_{-1}$, $X_{1} \rightarrow X_{0} \times _{X_{-1}} X_{0}, \ldots$

Is this functor $M$ a coCartesian fibration?

I tried to work out the easier question where we only look at $M_{0}: Fun(a \Delta^{op}, C) \rightarrow Fun(\Delta^{1}, C)$ which only remembers the first matching map. This is the same as the functor induced by the inclusion $\Delta^{1} \rightarrow a \Delta^{op}$ of $[-1], [0]$ into the simplex category, which I think (please correct me if I'm wrong) is left anodyne, being obtained by coning off a left anodyne $[0] \rightarrow \Delta^{op}$.

I'd actually would just want know if this $M$ is smooth (in the sense of HTT) but not sure how to show that without coCartesianness (4.1.2.15 gives the implication).

Hey @RuneHaugseng, I was wondering if your paper on the mate correspondence can possibly prove the following: given a space $X$ then we know the colimit functor $Fun(X,Sp)→Sp$ is colax monoidal (this follows immediately from your paper) with respect the pointwise monoidal structure. Then we know that, since any functor $f:X\to Sp$ is a comodule for the monoidal unit $\mathbb{S}_X$ (the constant functor) there is a coaction (by colax monoidality) $colim(f)\to colim(f)\otimes \mathbb{S}[X]$...

Of course this coaction is NOT the colimit of the coaction $f\to f\otimes\mathbb{S}_X$ in $Fun(X,Sp)$ (since that map is an equivalence, for one). But I'd like to identify it with the colimit of the unit map $f\to p^*p_!(f)$, where $p:X\to \ast$ is the terminal map.

The colimit of that map, if I'm not mistaken, will be a map $colim(f)\to colim(f)\otimes X_+$, and I feel like they should be the same map?

So I suppose I'm wondering if the "colax monoidality" of the colimit can be described really explicitly, at least in this case, to tell me exactly what the induced coaction map is.

Oh, in one place I've written $colim(f)\otimes \mathbb{S}[X]$ and in another I've written $colim(f)\otimes X_+$, but of course these are the same thing.

Maybe this will require cracking open the explicit dual fibration here described by Clark, Saul and Denis' paper.

Here's a modest version of the question about the Tate Frobenius for odd primes from a while back: Can we at least be confident that the iteration of the $\mathbb{E}^{\infty}$ Tate frobenius $A \to A^{t \Sigma_p \times \Sigma_p}$ is not symmetric w.r.t. permuting the 2 factors of $\Sigma_p$? (or maybe better that this sentence doesn't even make any sense etc...).

Small reminder: from the computation I tried to do this version of symmetry gave me the Adem relations (for the reduced powers - no bocksteins involved) without any signs. So either i messed up or this should be wrong so it's at least a well defined contradiction.

@SaalHardali certainly the iterated frobenius landing in (A^{t\Sigma_p})^{t\Sigma_p} is invariant under swapping the factors of \Sigma_p, by the same argument (since that's given by an inner automorphism in \Sigma_{p^2}). But my claim is that this will not be enough to conclude the Adem relations. Instead, one will need to use that the iterated Frobenius A-->(A^{tC_p})^{tC_p} is invariant under GL_2(Z/p) (or the corresponding fact for \Sigma_p).

That's weird. I guess I don't understand the argument even in the p=2 case. If I just write $Q_t x = \Sigma Q^j(x)\beta (t)^j + \Sigma \beta Q^j(x) \beta (t)^{j-1} t$ where $t \in H^{2p-3}(\Sigma_p;\mathbb{F}_p)$ is the generator. Then apply it twice and use cartan formula and that $Q_s(\beta(t)) = \beta(t)\beta(s)^{1-p}(\beta(s)+\beta(t))^{p-1}$ to get a series in 2 variables $\beta(s)$ and $\beta(t)$ which by the symmetry above must be symmetric for interchanging.

This is what gives me the adem relations with no signs

maybe i'm wrong to assume that the contribution from coefficient of $st$ in the iterated frobenius decouples from the contribution of the coefficient of 1 (i'm thinking of everything as a module over the even part which is $\mathbb{F}_p[[\beta(t),\beta(s)]][\beta(t)^{-1},\beta(s)^{-1}]$)

sorry if this is messy, it's hard to describe this computation without actually carrying it out

what i'm trying to say is that somehow just using the interchange i already manage to get a contradiction with the Adem relations. Since I'm using my understanding from the prime 2 i must be misunderstanding something already there...

maybe actually all the headache comes from the $C_p \neq \Sigma_p$ as you are hinting but I would still like to know why my naive approach fails...

maybe the tate frobenius power with the symmetric group is not a commutative ring?

haha

that would solve it

i just now realized this might be an issue

is the tate power $X \mapsto (X^{\otimes p})^{t G}$ functor lax symmetric monoidal for any transitive group $G \le \Sigma_p$?

you're interested in (-)^{\tau G} not (-)^{tG}, but either way yes that's true.

oh yeah by $t$ i should mean the proper tate if that's what $\tau$ stands for

yeah bad choice of notation by me

well it's not 'proper' tate

tate for the transitive family

right?

it's Tate wrt the family of subgroups which don't act transitively

yeah

but I thought it gives the same answer in the mod p situation

the proper and the family of transitive subgroups\

actually nevermind

basically i'm content with using proposition A.18 from arxiv.org/pdf/2002.02035.pdf

I guess I'm surprised no signs appear when you're computing Q(s)Q(t)... like, you really never move a t or an s past something else in that computation?

maybe i'm doing something wrong

the other possibility is that the coefficients in the power series differ from the classically defined operations by some signs

but in the way i've written it there's the coefficient of 1 and the coefficient of t

when you apply twice you have 4 series for 1,t,s, and ts

after all- the argument comparing those involves various features like the boundary map in some triangle... moreover, if you use \Sigma_p, then usually the cohomology of \Sigma_p is split off from that of C_p using some idempotent that has some sign in it, by convention, if I remember right... so there are a few places signs could appear

yeah there's surely a sign somewhere that i missing

but i don't think it comes from interchanging s and t

at least not in the way i set up the computation

it would come up when you want an adem relation for stuff like $\beta Q^a \beta Q^b$

but for $Q^a Q^b$ it doesn't pop up

@DylanWilson I haven't considered that possibility!

it is reasonable considering i've never seen a claim that these are actually the classically defined dayer lashof operations

but i'm not that familiar with that kind of literature so it might not mean much

I guess maybe it'd be weird if there were a sign at that step, since Nikolaus-Scholze show that F_p--->(F_p^{tC_p})hF_p^{\times} gives the usual P^i with no sign

oh i see

so maybe it's the idempotent thing

maybe my computation of the dayer lashof action on $\mathbb{F}_p^{\tau \Sigma_p}$ is wrong...

i just used that $\mathbb{F}^{\tau \Sigma_p} \to \mathbb{F}_p^{t C_p}$ is injective on homotopy

maybe that's a problem because this map is not a map of algebras?

or maybe it is but it doesn't induce the obvious invlclusion but rather upto signs? (that's maybe what you meant by the idempotent...)

Anyway this shit needs to be sorted out at some point, it's driving me crazy

sorry for the french ^^

sorry I dunno :( can't do signs... I come from a p=2 tradition

it wasn't a complaint to you, rather to the universe.

you're only helping here

thanks a ton!!!

Guess i'll need a signs expert to resolve this...

Oh by the way notational question, i've seen the dayer lashof operations in the odd prime case denote by $Q^j, \beta Q^j$ and by $P^j , \beta P^j$, what's the more conventional notation?

I've always seen P for the Steenrod operations and Q for the Dyer-Lashof operations

(yeah, I think P^i = Q^{-i}). anyway- I think the thing I find more surprising than signs is that you're getting like any information at all using swap instead of the interesting element of GL_2(Z/p)... it seems to me like just using swap symmetry gives nothing resembling the Adem relations (with or without signs). can you explain that step?

like, in the classical setting with spaces, if we just look at H^*(X)-->H^*(X\times BC_p) and iterate (so nothing with Tate or anything fancy) then we have swap symmetry but you don't really learn anything from it.

Then when you open the $Q_s$ you get 4 terms. When you interchange $t$ and $s$ you get the following:

$$\sum (Q^i Q^j x) \beta(s)^{i-j(p-1)} \beta(t)^j (\beta(t)+\beta(s))^{j(p-1)} = \sum (Q^i Q^j x) \beta(t)^{i-j(p-1)}\beta(s)^j (\beta(t)+\beta(s))^{j(p-1)}$$

For the terms that do not involve the $\beta Q^j$'s

Then I used the method of residues to get the adem relations from this

When I do this i get the exact formula for the adem coefficients but with no signs

By taking the coefficient of $u^j$ on both sides where $u = \beta(s)^{1-p}\beta(t) (\beta(t) + \beta(s))^{p-1}$

@DylanWilson I hope any typos i might have made do not obscure the general idea of what i did

@SaalHardali in the mod 2 case, when I do the residue calculation, I use that 'du = ds' (or maybe you've got it set up so that du = dt or something); but that doesn't look true in this case. if you think of u, in your setup, as a power series in \beta(t) with coefficients in k((\beta(s))), then du is something more complicated

writing x=\beta(s) and y=\beta(t), I get something like x^{1-p}(x+y)^{p-2}[(x+y)^{1-p}-y] dy (but presumably there's at least a few mistakes because I didn't keep doing the computation until my answer stabilized...)

anyway- did that enter into your calculation?

[also, maybe we should move this conversation to email since we're way in the weeds now...]

I did not use $du =ds$ but rather i computed $du$ and made the necessary adjustments. Of course i might have made a mistake there. And yes maybe it's better to continue this somewhere else...

I'm not 100% sure this question makes sense but... is a limit of Cartesian lifts a Cartesian lift?

Like I've got a Cartesian fibration $p:E\to B$ and some map $h:b\to b'$ in the base. I've got a diagram of things living over $b'$, let's call them $e'$ and this induces, by taking Cartesian lifts, a diagram of morphisms over $h$ I think, let's say $f:e\to e'$. Then by taking limits I get a diagram $lim(h):lim(e)\to lim(e')$ right? But is this the Cartesian lift of $h$ with target $lim(e')$?

It seems like this should follow from the universal property of the Cartesian lifts?

Like, they satisfy the same universal property, sort of?