@Dedalus If you define the twisted arrow category as a Segal space (so Tw(C) is the simplicial space Map([n]*[n]^op, C)) then the 0th space is by definition Map([1], C), so I guess you can prove this by checking this Segal space is complete?

@MaximeRamzi oh cool, which paper?

@SaalHardali would you mind elaborating on this? it sounds like you are asserting that this is trivial e.g. if $\pi_1(Z)$ is trivial? i'm not even aware of how to make the argument in this case.

ohhh should i be thinking of $Z=BG$ for some $\infty$-group $G$, and $X \to Z \leftarrow Y$ as $M_{hG} \to BG \leftarrow N_{hG}$? then, the fiber product is $(M \times N)_{hG}$. before taking homotopy quotients, i just have a product square for $M \times N$. i can map this into my (commutative ring) spectrum $A$, and then apply the exact (and lax symmetric monoidal) functor $(-)^{hG}$. so perhaps this just boils down to understanding maps out of a product of spaces.

Sorry I should have replaced $Sp$ everywhere with $Mod_R$

The way i've written it $R = \mathbb{S}$

The spectral sequence obtained from the first of these cosimplicial diagrams is exactly the eilenberg moore spectral sequence for homology (whose dual is the EMSS for cohomology).

So of course there are many nitpicks like convergence and taking dual issues. But that's the way i like to think about it.

Then you observe that this is trivial for the trivial local system on $B$. And that this property is stable under finite limits/colimits (because everything is stable).

So that way you get it for all maps whose pushforward is a "nilpotent" as local system of $R$-modules.

Coming back to what I said earlier I would love to see an example where the EMSS converges for reasons that can't be traced back to this fact.

oh sorry, I should probably mention that this box tensor product should stand for co-tensor product.

There is also a discussion on convergence of the EMSS in Section 7.2 of arxiv.org/pdf/1507.06869.pdf

@AaronMazel-Gee The paper is "Le problème de la schématisation de Grothendieck revisité." - it's in french, but if you're interested I wrote up a slightly different version of the argument (he uses topological spaces to model the homotopy pullback, which I wasn't happy about, so I rewrote it with homotopy types; and also in english - but it's really essentially his argument)

As Saal points out, in the end you're bootstrapping your way up from the trivial system by using a nilpotence hypothesis (which you can read as "being in the thick subcategory generated by the trivial system")

@Drew Yes, although I think it's only in the case of the "path fibration" pullback

Suppose I have a braided monoidal ∞-category C. Then both LMod(C) and Alg(C) inherit monoidal structures from C. Suppose I have an algebra object in C (A,M). I think this is going to mean that, in particular, M is an algebra in C. Is there some way to see that M is an algebra in LMod_A(C) as well?

Or some way to see why this might not be true? Or conditions to make it true?

Sorry, also A should be an $\mathbb{E}_2$-algebra.

So I get a monoidal structure on LMod_A(C) by pulling back the cocartesian fibration LMod(C)^⊗→Alg(C)^⊗ along the map Ass^⊗→Alg(C)^⊗ picking out A as an $\mathbb{E}_2$-algebra.

(this is of course assuming that tensor product commutes with geometric realization in C)

I'm sort of thinking it might come from a diagram like this:

Your rightmost Ass could map to the bottom Ass as well, making the whole thing commute; so it factors through the pullback, doesn't it ?

(forgetting about the Ass x_{Alg} Ass, I'm talking about the lower pullback)

Righhhhttt.... okay yeah so we get a commutative triangle Alg(C)^⊗←Ass^⊗→LMod(C)^⊗→Alg(C)^⊗?

Yup, but you can see it as a square if you put two Ass's in there

God I should not have used "Ass"

(which paper was it in which the authors said they changed the operad's name for that reason ?)

Haha, I don't know. Doesn't Lurie write Assoc now?

"A sock"

I think so, in my version of HA yeah

Okay but anyway, right, I think I'm now overcomplicating it but the point is we get the following diagram:

Which picks out M as an algebra in LMod_A(C)

I'm anxious I need to be careful about things like "pullbacks of ∞-operads?"

Well I definitely don't know the references about that, but along a strict monoidal functor which is also a fibration, it should work out fine

Hrm... I don't think either of the legs of the pullback here are strict monoidal?

LMod(C) -> Alg(C) is

but I was thinking of the smaller one

Right. The bottom square is probably fine.

and it's enough to get what you want

Oh okay so maybe I do need to figure out what you were saying about making a smaller square, haha.

Oh I'm a doof. I see what you were saying.

Ah cool !

yes exactly !

Thanks very much @MaximeRamzi that puts my mind at ease very much

My pleasure :)

I'm confused about something different : it's well known that 2 isn't nullhomotopic on $S^0/2$ (for instance one of the homotopy groups has $4$-torsion). However the following argument seems to show that it does : Consider $S^0$ as a $S^0[t]$-module (the free $E_1$-ring on a generator in degree $0$, which I am seeing as an $E_\infty$-ring) via the $0$-endomorphism. Then the action of $t$ is equivalent, as a $S^0[t]$-module map, to $0$.

Consider a second sphere, on which $S^0[t]$ acts as $2$ which I'm gonnna call $X$ to avoid confusion. Then $X\otimes_{S^0[t]}S^0$ is equivalent to $S^0/2$. On it, $2$ acts as the $2$ on $X$ which is the action of $t$ on $X$, and so it acts as $t$ on $S^0$, i.e. as $0$. So $2$ is $0$ on $S^0/2$ : what ?

To prove that $t$ acts as $0$ on $S^0$, $S^0[t]$-linearly, and as $2$ on $X$, $S^0[t]$-linearly, one can use the $E_1$-morphisms $S^0[t] \to S^0$ induced by $t\mapsto 0$ and $t\mapsto 2$ respectively

:57157133 There's an $E_1$-morphism $t\mapsto 2$ along which I can pullback

Got it.

But now I realize that it may actually be in that last step: $S^0[t]\to S^0$ is only $E_1$ (is it ?) so I can't pullback the action of the ring along it, can I ?

So $t\mapsto 0$ is definitely $E_\infty$ (it's induced by $\mathbb N\to *$), but maybe $t\mapsto 2$ isn't ?

and so actually, $t$ doesn't act as $2$ on $X$, $S^0[t]$-linearly. Is that what it is ?

that would be freaky

Yes, $t\mapsto 2$ is not $E_\infty$.

@CharlesRezk Is there an easy way to see that ?

If it were, then $X\otimes_{S[t]}S$ would be an $E_\infty$-ring $S/2$.

Ah, of course, duh

Cause it's a tensor product of two $E_\infty$ algebras.

I guess.

Yes.

Ok, is there an easier way to see that $t$ doesn't act on $X$ as $2$, $S^0[t]$-linearly, or is my confusion the simplest ?

@CharlesRezk (thanks by the way !!)

I guess it also provides an example of an $E_1$-ring $A\to B$, where $A$ "is" $E_\infty$ such that $B$ is not an $E_1$-algebra in $A$-modules !

This should read $E_1$-ring map*

So I guess this is why it works well over $\mathbb Z$ : $t\mapsto 2$ "is" $E_\infty$ as a morphism $\mathbb Z[t]\to \mathbb Z$

Is that what people call strictly commuting elements ? And so $2$ isn't strictly commuting in $S^0$ ? For some reason this strikes me as weirder than "$2\neq 0$ on $S^0/2$" but it probably shouldn't

@MaximeRamzi for what it's worth, the statement about pullbacks of ∞-operads being well behaved in this situation follows from HA.2.1.4.6 combined with the fact that cocartesian fibration are categorical fibrations and therefore fibrations in the model structure on ∞-operads, and ∞-operads are fibrant.

ah, that's good to know, thanks !

@MaximeRamzi I think one good way to see that 2 doesn't act by 0 on $S^0/2$ is that the cofiber of 2 is $(S^0/2)^{\otimes 2}$ and if 2 did act by zero then the cofiber would be $S^0/2 \oplus \Sigma S^/2$; these have inequivalent mod 2 cohomology over the steenrod algebra.

Gah, meant to type $S^{0}/2 \oplus \Sigma S^{0}/2$

Coincidentally, this is the same computation which you can use to prove the mod 2 Moore spectrum fails to admit a unital multiplication.

@LiamKeenan right but I guess I got confused (and actually still am) about why $t$ doesn't act as $2$ $S^0[t]$-linearly on $S^0$

even though it is defined to act that way $S^0$-linearly

(the reason I'm confused about it that when I look at $Mod_{S^0[t]}$ as $Fun(\Delta^1/\partial \Delta^1, Sp)$, the statement seems very innocent

is*

but I feel like I get it now, by writing things in terms of $B\mathbb N$ actions on oo-categories

Does anyone know of a characterization of the fibrations in the cocartesian model structure on $(sSet^+)_{/S}$?

Is 2 nilpotent on S/2?

@WilliamBalderrama Yeah it squares to 0; in fact for any stable oo-cat C and any self map f : X -> X, the induced map X/f -> X/f (call it f if you like) is not always 0 but it always squares to 0

I see.

@MaximeRamzi Strictly commuting elements are pretty rare. In general I'm more surprised to learn an element is strictly commuting than to learn it isn't.

One way to see that 2 isn't strictly commuting is with power operations. If R is E-infinity, then there's an external squaring operation P : pi_0 R -> R^0 BSigma_2 satisfying the Cartan formula P(x+y) = P(x) + P(y) + transfer(xy).

If x is strictly commuting, then P(x) is just supported on the basepoint, i.e. "P(x) = x^2". But P(2) = 2 + transfer(1) is generally nonzero.

Err, "is generally nonzero" -> "is generally different from 4".

@WilliamBalderrama Ah thanks, that's a good point of view too; it feels like it's related to what Liam was saying with the Steenrod algebra, but I know too little about power operations to actually come up with the connection without thinking

And I guess I knew that strictly commuting elements were pretty rare, but for some reason the integers seemed so "basic" that they had to be. But of course if you think about what 2 is in the sphere, it's closer to like the finite set 2 than the integer 2, so the twist 2 x 2 -> 2 x 2 has no reason to be trivial; which I guess is what strict commutativity would be saying

For some time I had the impression that $Q^1(h_0)=h_1$ on the E_2 page of the ASS implies $Q^1(2)=\eta$ in $\pi_*(S)$. I never went carefully through this though...

Essentially the question is under what conditions (maybe none) the algebraic power operations match with the topological ones.

Of course my question shouldn't be interpreted in the sense of "all true statements are equivalent" haha

To be honest I have the same question about massey products. When do the algebraic ones match with the topological ones...

Maybe the thing to say is it always matches upto extension problems and then you just check that $\eta$ is the only class in $\pi_1$ so there are no lower filtra corrections.