7:53 AM
@FrankScience I have been confused about several related things in the past -- however, Nikolaus-Scholze seem to indicate that the Tate diagonal is lax symmetric monoidal. where have you found that these don't commute?
regarding cosimplicial commutative things vs filtered commutative things, the two are indeed different
here's one example. if you have a commutative monoid in filtered objects, then the E_1-page of the associated spectral sequence is graded-commutative
however, the same is not true for cosimplicial objects.
e.g.: say A is a commutative Hopf algebra over k -- its cobar complex C(k,A,k) is then a cosimplicial commutative ring. in the E_1-term, the 1-line consists of equivalence classes [x] where x is primitive
the product of [x] and [y] is [x|y] in the cobar complex, which is not the same as [y|x]. the two become equal at E_2 because their difference is the boundary of [xy]
(oops. got the sign wrong)
more generally, if you have a commutative filtered object ... -> A_2 -> A_1 -> A_0, the Day convolution gives you, e.g., (A_n ⊗ A_n) / Sigma_2 -> A_{2n}.
the same isn't true if you take a cosimplicial commutative object and convert it into a filtered object.
you get a map (A_n⊗A_n)/Sigma_2 -> A_n. it so happens that the map A_n⊗A_n -> A_n has a lift to A_{2n} -- but not the whole homotopy orbit.
in general the map from the homotopy orbit is kind of "spread out" between different layers of the filtration between n and 2n.
this isn't an "oh, things are better if you work coherently" thing -- it's fundamental to the Dold-Kan correspondence.
now, having said that, it's easy to get confused here because sometimes people conflate two different situations
the classical Dold-Kan correspondence is two things
first and foremost, it is an equivalence of categories between nonnegative cochain complexes and cosimplicial objects
second, it is also an equivalence of homotopy theories between (nonnegative cochain complexes + quasi-isos) and (cosimplicial objects + weak equivs)
in the first case, the dold-kan correspondence is lax symmetric monoidal in one direction (using the eilenberg-zilber shuffle map) and only lax monoidal in the other (using the alexander-whitney map), full stop
when you add in the homotopy theory, the eilenberg-zilber map becomes a natural weak equivalence. so you get a symmetric monoidal equivalence of homotopy theories between cochain complexes and cosimplicial abelian groups.
but the high-powered version of the Dold-Kan correspondence for stable categories isn't a generalization of the second case. it's a generalization of the first case
(meaning, e.g., you have an equivalence between cosimplicial spectra and filtered spectra, but on both sides the equivalences are defined objectwise, not some extra structure analogous to weak equivalences of complexes)
Alice Hedenlund has some notes on the filtered approach to spectral sequences here: math.ru.nl/~sagave/east2018/tate_construction_talk_EAST2018.pdf
those notes indicate forthcoming work about multiplicativity.
for the record:
(a) the lax-symmetric-monoidal eilenberg-zilber map is known to lift, so that every commutative object in filtered spectra becomes a commutative object in cosimplicial spectra.
(b) i am not aware of anybody who has managed to get the alexander-whitney map to lift to the coherent setting, which is what you would need to assert that the other half of D-K is lax monoidal. i thought i had an argument once, but it foundered on my lack of ability to make the technical details go (eg I don't know how to make "Stasheff associahedra" arguments for imposing associative algebra structures in the new framework)
(c) random, related: why didn't anybody ever tell me that the alexander-whitney map is geometric? if you take the subspace of (Δ^n x Δ^n) consisting of the union of (front p-face) x (back q-face), that's a subspace homeomorphic to Δ^n
i didn't know that until six months ago
(d) in some sense this is why multiplicativity of things like the Serre spectral sequence can be such a pain in the neck to prove, it's because the commutativity isn't on-the-nose on the filtered level
(e) way back up there, the elements on the E_1-page of a cobar complex are general elements [x], and the cocycles are the ones where x is primitive. sorry

4 hours later…
12:13 PM
@TylerLawson FWIW Stasheff associahedra are essentially using the standard argument to extend a section of a planar operad from $Δ^{op}_{\le n-1}$ to $Δ^{op}_{\le n}$ by looking at $Δ^{op}_{\le n-1}×_Δ Δ_{/[n]}$ (the "boundary" of the Stasheff associahedron)
(here I'm thinking of an associative algebra as a section of a map C^⊗→Δ^{op} sending inert arrows to inert arrows)

1:02 PM
@TylerLawson Thanks very much for clarifying these things! I had a "program" to think about this problem, but never concentrated on it too much. My first sanity check would have been to show that cosimplicial -> filtered is lax monoidal - but I did not try this either.
I wonder if one can make sense of an operad E^fil (in the classiacl sense) valued in filtered spaces, where E^fil(n) is the space E Sigma_n, with its usual skeletal filtration. Then one should take algebras in filtered spectra for this operad, call the category FAlg. (1) Does CAlg(cosimplicial) map naturally to FAlg? (2) Can FAlg be described more intrinsically?
(These guesses are based on Bruner's work on multiplicative structures in the adams tower.)
When I was looking at this it seemed like in order for this to work it may be necessary to localize filtered spectra at the "E^1-equivalences", i.e. maps inducing equivalences on the cofibers X_n -> X_{n+1} for each n. But I'm not sure.

@DenisNardin right, so: let's call Delta_{/[n]} = K (the "stasheff associahedron") and its boudary B; you get a map from the pushout of K <- B -> Delta^op_{\leq n-1} to Delta^op_{\leq n}.
Is it straightforward to show that map is marked anodyne?
@TomBachmann cosimplicial -> filtered being lax monoidal should be lax monoidal, but it's the tough direction for sure
It seems reasonable that there's some kind of way to say it using filtrations like that -- I'm not sure.

1:25 PM
@TylerLawson I have noticed that on BMS18, they exhibited a cyclotomic structure on $\DeclareMathOperator\THH{THH}\THH(A/\mathbb S[z])$ when A is a (commutative, I don't know whether this is necessary) $\mathbb S[z]$-algebra, where $\mathbb S[z]$ is the free $\mathbb E_\infty$-algebra with a single generator $z$.
Construction 11.5
It seems to me that the key point is that the projection $\THH(\mathbb S[z])\to\mathbb S[z]$ commutes with Frobenii (Proposition 11.3)
The projection $\THH(A)\to A$ looks like a multiplication $A^{\otimes m}\to A$.
As far as I understood, the Frobenius on $\THH(A)$ is induced by Tate diagonals $A^{\otimes m}\to(A^{\otimes mp})^{tC_p}$.

@TylerLawson I think it works better by taking K=(Δ_{\le n})_{/[n]}? I don't think it is straightforward, but I think it is the same sort of argument that shows it is a categorical equivalence (as in Bousfield-Kan). I've tried to write it down for a couple of minutes and I failed, though, so it might harder than I thought (Lurie does talk about the associahedron in HA, but he uses a completely different method)
@FrankScience What's the projection $THH(A)→A$? I thought that the canonical map went in the other direction

Ah, sorry, $A$ should be commutative, otherwise no multiplication would be available.

Sure, but still, how do you define the thing?
Oh you do the multiplication map levelwise, got it

I believe that it is, although in Nikolaus-Scholze, they produce it from the universal property.

I think that the Tate-level Frobenius does commute with multiplication, since it is a map of rings. The Frobenius on THH is slightly subtler (you can't just say "Tate diagonal levelwise")

1:34 PM
They claim that $A\to\THH(A)$ is initial among $S^1$-equivariant commutative algebras.

Right, $THH(A)=A^{⊗S¹}$
Oh and you're taking the map $S^1→*$

Then the identity $A\to A$ induces the map in question. I think that it coincides with the multiplication map.

Yeah, I see why: the multiplication map is just induced by the map $n→*$, so it can be factored $n→S^1→*$ where the first map is just embedding n points into $S^1$
So, I'm not an expert, but it seems to me that the confusion might come from the fact that we are using "Frobenius" to describe two a priori different things
One is just the canonical map $A→A^{tC_p}$ that every commutative ring has.
The other is the structure map in a cyclotomic structure
I think (but I'm not 100% sure) that in proposition 11.3 the key point is that the top arrow is the latter, while the bottom arrow is the former

They are closely related.

Indeed, but not exactly the same, so you need a small proof for the compatibility result
Also, $\mathbb{S}[z]$ is not the free commutative ring on one generator (this one is a nitpick, but since I'm here I might as well say it)

1:41 PM
You are correct
$\mathbb S[z]$ is not the free commutative ring.
it is "very commutative"
I mean
The commutative diagram in Proposition 11.3 seems to be the essential thing that does not work for general $R$, only for $\mathbb S[z]$ (I think that it also works for $\mathbb S[M]$ where $M$ is a commutative monoid)
I am trying to understand where it does not work for a general $R$.
Let me fix the notation. Denote by $R$ the "base" commutative ring and by $A$ an $R$-algebra.

It would be nice to have a counterexample for a general $R$. Have you tried $R=\mathbb{F}_p$?
(also, I've got to run away, sorry to leave you suddenly)

There is no cyclotomic structure on $\THH(A/R)$ in general, say, $R=\mathbb F_p$ or $\mathbb Z$, etc.

Sure, I was saying for the analog of proposition 11.3

As far as I understand from 11.6, it seems to me that this commutative diagram is the key point to furnish a cyclotomic structure. If you had a diagram from $R$, then you would have a cyclotomic structure for $\THH(A/R)\cong\THH(A)\otimes_{\THH(R)}R$ by tracing their proof.

Yes, I agree

1:54 PM
Well, and I run away and I will be back in an hour :)

@FrankScience this is a point that I have been confused on myself in the past (to the point of making public errors when I state things).
here is what I think the problem is:
- the Tate diagonal X -> (X⊗X⊗...⊗X)^{tC_p} is lax symmetric monoidal: true
- the Tate construction, and the Tate diagonal, are also invariant under the C_p-action on X⊗X⊗...⊗X: true
- these two "invariances" for the Tate diagonal commute with each other: false.
e.g., the multiplication (X⊗X)^{tC_2} ⊗ (Y⊗Y)^{tC_2} -> (X⊗Y⊗X⊗Y)^{tC_2} requires knowledge of which factor of X gets multiplied by which factor of Y.
this makes the square involving the maps THH(A) -> A -> A^{tC_p} and THH(A) -> THH(A)^{tC_p} -> A^{tC_p} not commute in general: if you try to make it so, then you run into a problem with compatibility of the different face maps that are happening in simplicial degree 1.
i shouldn't use the past tense. this still regularly confuses me a lot
@DenisNardin I did not realize that this section on the associahedron was in here

2 hours later…
3:45 PM
@TylerLawson It seems that what you said also applies to the space level version - which I thought was correct. Given a commutative monoid M, we consider two compositions $\DeclareMathOperator\Bcyc{B^{\mathrm cyc}}\Bcyc M\to M\to M^{hC_p}$ and $\Bcyc M\to(\Bcyc M)^{hC_p}\to M^{hC_p}$ which don't coincide for general commutative monoids $M$?
I thought that $M^{\times n}\to M\to M^{hC_p}$ coincides with $M^{\times n}\to(M^{\times np})^{hC_p}\to M^{hC_p}$ and take colim over $n\in\Delta^{\operatorname{op}}$.

3 hours later…
6:29 PM
@FrankScience If it's a genuinely commuting monoid, I believe that you are OK because you can factor through fixed elements, rather than homotopy fixed elements. But I think that you are correct in the same objection does apply to E_infty spaces.
e.g.: let's say C is a symmetric monoidal category. if we have a map B^{cyc}(C) -> D of categories, we in particular have a functor F: C -> D together with a natural isomorphism T: F(A⊗B) -> F(B⊗A). it factors through the augmentation B^{cyc}(C) -> C if and only if T is F(twist).
now, the Tate diagonal B^{cyc}(C) -> B^{cyc}(C)^{hC_2} -> C^{hC_2} has the structure of such a functor.
(namely, via the natural transformation of diagrams that gives rise to the Tate diagonal)
if we trace it explicitly, we have C x C -> (C^4)^{hC_2}, then the two functors (C^4)^{hC_2} -> (C^2)^{hC_2}, and then the multiplication (C^2)^{hC_2} -> C^{hC_2}.
this composite sends (A,B) first to (A,B,A,B) with the permutation isomorphism; then the two functors send it to (A⊗B,A⊗B) with the flip iso and (B⊗A, B⊗A) with the flip iso; then it maps to A⊗B⊗A⊗B, and the natural isomorphism T is the two-fold twist.
my apologies, no. the two images are A⊗B⊗A⊗B and B⊗A⊗B⊗A, and the natural isomorphism is the cyclic permutation.
however, under the other composite C^2 -> C -> (C^2)^{hC_2} -> C^{hC_2}, the twist isomorphism does not go to this cyclic permutation.

6:46 PM
I am super confused. What is the essential difference between stable and unstable versions?

I'm claiming that the essential difference isn't between stable and unstable versions, but between coherent commutativity and strict commutativity (which we don't really have, stably)
I'm trying (but doing poorly) to say that the fact that you can't make the augmentation THH(A) -> A compatible with the Tate diagonal / Tate frobenius is a failure of coherence homotopies (or natural isomorphisms) to assemble, even though everything commutes in the homotopy category.

7:09 PM
Thanks. This seems a difficult task.

I'd be happy if I could just stop making mistakes about it, tbh

A formal proof of impossibility would extract some algebraic invariants - I saw proofs on papers like this, but I don't know how these invariants are usually extracted.