@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?
@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.
@AaronMazel-Gee Let $E \to B$ be a map and $R$ be a ring spectrum. Then asking for EMSS for $R$-homologyof the pullback along any $X \to B$ to converge is more or less equivalent to the beck chevalley property for the following diagram:
This is because on the one hand: $$\Sigma^{\infty}_+ X \boxtimes_{\Sigma^{\infty}_+ B} \Sigma^{\infty}_+ E = Tot (Cobar^{\bullet}(\Sigma^{\infty}_+ X,\Sigma^{\infty}_+ B,\Sigma^{\infty}_+ E) ) = Tot(\Sigma^{\infty}_+ Cobar^{\bullet}(X,B,E) ) $$
And on the other hand: $$\Sigma^{\infty}_+ f^{\ast}X = \Sigma^{\infty}_+ (X \times_B E) = \Sigma^{\infty}_+ Tot(Cobar^{\bullet}(X,B,E))$$
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.
@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:
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
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 ?
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.
@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.
@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
@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.
@JonathanBeardsley The fibrations between fibrant objects are the maps of marked simplicial sets over $S$ satisfying the op-duals of the conditions in HTT.3.1.1.6. You shouldn't expect any intrinsic characterization of the fibrations between arbitrary objects (consider $S = \Delta[0]$).