@RuneHaugseng How do you "flip the handedness in the first coordinate"? I think that was the fundamental issue I ran into when I tried to do this.

@JonathanBeardsley an $\infty$-cosmos is an $(\infty,2)$-category with all finite limits. You shouldn't expect an $\infty$-cosmos to contain nerves of ordinary categories, in general. But you can just say "let $\mathcal E$ be an $\infty$-cosmos equipped with a fully faithful functor of $\infty$-cosmoi $N\colon \mathrm{Cat} \to \mathcal E$..."

Does it ever make sense to take the fibre product of single-colored operads?

(or globular operads, which is the real question)

Is being MU acyclic equivalent to being K(n)-acyclic for all n including infinity?

@dhy You can view the map (say) E -> Delta^op x Delta^1 as a morphism of cocartesian fibrations over Delta^op which preserves cocartesian morphisms (this just says cocartesian morphisms over Delta^op in E lie over identities in Delta^1). So this corresponds to a morphism between the dual cartesian fibrations E' -> Delta x Delta^1.

Actually it's not obvious the composite E -> Delta^op is cocartesian...

Hrmm.. isn't this exactly what homology groups are supposed to capture?

It's not clear to me why you'd expect the torus to have only one "hole" though

Well then, that's an answer. But the number of holes you get might not agree with your intuition

For example $S^2$ is simply connected but it has a "2-dimensional hole"

Its simple connectedness is reflected in the fact that it has no "1-dimensional holes'

Well, to be a bit more precise, for every $n$ there is a definition of a group $H_n(X)$

I'm using the rank of that group as a proxy for the "number of holes"

It's totally unclear to me why you'd want to "free the definition from the restriction of dimension"

One clue that this might match your intuition is that if your $X$ is nice (locally contractible) and compact and you embed it in $\mathbb{R}^N$, the number of bounded components of the complement is exactly the rank of $H_{N-1}(X)$

I learned homology as an undergrad. It's not that hard. The intuition might take a while to develop though

@RuneHaugseng But can you describe this construction at the level of simplicial sets somehow? I couldn't figure out how to explicitize it enough to actually prove anything

@dhy If I got it right, it follows from a pretty simple observation about fibrations. I wrote up a short proof here: dropbox.com/s/emln6asthahbwps/oplaxladj.pdf?dl=1

@RuneHaugseng You should clean it up a little bit and put it on the arxiv, so people have something to cite

@SaalHardali I don't believe so. In particular, it is not the case that $\langle MU\rangle = \vee_{0\leq n\leq\infty} \langle K(n)\rangle$

However, I seem to recall that this is the case if $X$ is a finite cell complex.

@JonathanBeardsley For X finite in both cases acyclic implies 0. Do you have a counter example or a reference?

Well, the fact that $\langle MU\rangle \neq \vee_{0\leq n\leq \infty}\langle K(n)\rangle$ is in Ravenel's '84 paper (web.math.rochester.edu/people/faculty/doug/mypapers/loc.pdf) at least if you replace $MU$ with $BP$

(Section 3, in particular)

Erm, no not Section 3... where is it...

I believe the spectra $BPJ$ of definition 2.7 have Bousfield classes in between $BP$ and the wedge of all Morava $K$-theories.

@JonathanBeardsley Thanks, ill make sure to check there.

@SaalHardali @JonathanBeardsley Being $\vee_n K(n)$-local is called being harmonic, right? And I guess I thought that every spectrum was $MU$-local, but I suppose I'm probably just thinking of the finite case.

@TimCampion No, that is not true. It is true that if you have an $MU$-acyclic ring spectrum this is 0 (that's a reformulation of the nilpotence theorem), but the hypothesis "ring" there is essential

the anderson dual spectrum is not MU-local

if E is MU-acyclic, then it is necessarily K(n)-acyclic for all n geq 0. the converse is not true: HF_p is K(n)-acyclic for all n geq 0, but it is not MU-acyclic.

but if you include K(infty), that example fails because K(infty) is HF_p

in fact, i would believe that the bousfield class of MU is the same as the bousfield class of the wedge of K(n) for 0 leq n leq infty

@skd do you know a counter example when K(infty) is included?

no

Sorry for the odd question apparently I wrote it before I noticed you aaying that you think they are bousfield equivalent.

Do MU and HZ have the same Bousfield class, on account of being connective ring spectra with pi_0 = Z? And then does HZ have the same Bousfield class as HQ v bigvee_p HFp = K(0) v bigvee_p K(∞)? If so, that would do it & would point out that for this question the intermediate K(n)s are not so relevant

But I fear that I am rusty & wrong

I guess that can't be: K(n) has nontrivial BP-homology but has trivial HZ-homology. Bummer

I think the theorem says that if X is connective, then it is both MU-local and E-local

(more generally, if X is connective then L_EX=L_{Hπ_0E}X for all connective E)

np, @SaalHardali

this is an interesting question

I get stuck with Remark 25.1.3.7 on Lurie's Spectral Algebraic Geometry. For sake of completeness, let's fix some notations.

Given a (discrete) commutative ring R and denote by $\operatorname{SCA}_R:=\operatorname{Fun}^{\pi}(\operatorname{Poly}_R^{\operator‌​name{op}},\mathcal S)$ the non-abelian derived category of the category $\operatorname{Poly}_R$ of $R$-polynomials.

This is the $\infty$-category of simplicial commutative algebras over $R$.

Now given a $R$-algebra morphism $f\colon R[x_1,\dots,x_n]\to R[y_1,\dots,y_m]$ given by $x_i\mapsto f_i(y_1,\dots,y_m)$ and an object $A\in\operatorname{SCA}_R$

We have a map of spaces induced by $f$: $A(R[y_1,\dots,y_m])\to A(R[x_1,\dots,x_n])$

In the previous remark, he identifies $A(R[x_1,\dots,x_n])$ with $A(R[X])^n$.

Now he claims that under this identification, taking the homotopy groups of the previously constructed map of spaces, we get a map $\pi_*(A)^m\to\pi_*(A)^n$, and for $*=0$, the map is given by $(a_1,\dots,a_m)\mapsto(f_1(a_1,\dots,a_m),\dots,f_n(a_1,\dots,a_m))$.

and for $*>0$ (and we assume that these spaces are pointed at $(a_1,\dots,a_m)$ and $(f_1(a),\dots,f_n(a))$), the map is given by the Jacobian matrix of $f$ at $(a_1,\dots,a_m)$.

[Note: You don't need to assume pointings, we are working in simplicial abelian groups where the basepoint is irrelevant]

I have no idea how these gadgets could be computed.

Actually it's even more concrete than that

For $*=0$, I can conceive that the ring structure of $\pi_*(A)$ is induced by $R[z]\to R[x,y],z\mapsto x+y$ and $z\mapsto xy$.

These, before being simplicial rings, are simplicial abelian groups, so you can compute the homotopy groups as the homology groups of the associated complex

I know this but I wonder whether I can compute stuff directly through the derived categorical definition.

So a class in π_nA is represented by an element x∊A_n such that d_ix=0 for all i

Uhm, then I think you have to go back to the universal example

You mean, for polynomials?

Well, I wish. The universal simplicial ring with a class in degree n is the symmetric algebra over $\tilde C_*(S^n)$

Note that as a simplicial abelian group $\tilde C_*(S^n)$ has tons of (degenerate) classes in higher degree

Frankly I don't see a proof of this formula using only abstract properties of the derived category

I wonder, say, if $A$ is a connective commutative ring spectrum over $R$ and $f\in R[X_1,\dots,X_n]$, do we have something similar like $(\pi_*A)^n\to\pi_*A$ induced by $f$?

Hah hold on. The object representing $(π_*A)^n$ is not $R[X_1,...,X_n]$ in this case

That's exactly the difference between ring spectra and simplicial rings

oh

So, yes, you do get a morphism $[R[X_1,...,X_n],A]→[R[X],A]$, but neither side is what you expect

Well, unless R is a Q-algebra :).

However now that I think about it... if we denote with $R\{X_1,...,X_n\}$ the free commutative R-algebra (in spectra) on n generators in degree 0, maybe $π_0R\{X_1,...,X_n\}=R[X_1,...,X_n]$ so we still get a formula like you want

This only works for $\pi_0$

Well, no. Once you have a map $R\{X\}→R\{X_1,...,X_n\}$ you're good to go

Of course there will be higher operations, but let's take this slowly :)

Anyway, it seems strange that one should go back to (classical) simplicial objects to obtain Lurie's remark above. It seems that this could be done merely by definition-tracing.

I don't feel like that. A completely abstract proof would work in any Lawvere theory (e.g. for example for simplicial Lie algebras, or simplicial abelian groups or...) but the formula we got says something specific about rings (i.e. that the operation in higher degree is controlled by the "first order terms")

There might be something general to say if you're clever, but at the end you're gonna have to do some kind of computation

I don't know the general theory, but for general one, you have things like the functor like $\operatorname{SCR}\to\operatorname{CAlg}(\operatorname{Sp})$?

Uh I'm talking about a rather different thing

Let me give you another example. Let $\Lambda$ be the category of free Lie algebras on a finite set and Lie algebra homomorphism

Then as before you can define the ∞-category $\mathrm{Fun}^{\times}(\Lambda^{op},\mathcal{S})$ of product preserving functors from $\Lambda^{op}$ to spaces

The general theory says that this is the ∞-category of simplicial Lie algebras (i.e. it is the ∞-category you obtain by taking the category of simplicial Lie algebras and inverting the maps that are weak equivalences of the underlying simplicial sets)

If $F(n)$ is the free Lie algebra on $n$-generators, the presheaf represented by $F(n)$ corepresents $L^n$

So for every element of $F(n)$ I get a map $(π_*L)^n→π_*L$ that in degree 0 is just given by the Lie algebra structure on $π_0L$

Whatever abstract trickery you find for the previous case should work here too. But I have no clue of what should replace the Jacobian

How does $\pi_0$ act on $\pi_*$?

Well π_*L inherits a graded Lie algebra structure (although don't ask me about signs)

So π_*L is in particular a π_0L-module (in the sense of the representation theory of Lie algebras)

But as you can see we can play this game with pretty much any algebraic theory: commutative rings, associative rings, quandles, Lie algebras, restricted Lie algebras... There's just no way there's a unique formula for all of them

Actually now that I write it I'm less convinced

Maybe we can be clever using derivations....

Anyway I think I should go to bed :)

I don't now how to do that.

No, me neither :)

In fact, in my example, I inherit the graded commutative structure of $\pi_*A$ from the associated ring spectrum.

I mean this simple statement that $\pi_*$ inherits the structure you start with.

Well, π_*A has more structure. You get a full divided power structure with respect to the ideal of elements of positive degree

That's not coming from the associated ring spectrum

@SaalHardali @JonathanBeardsley @EricPeterson @DenisNardin and anyone else who's interested: tobias barthel pointed out to me that the proof of 2.2 in Ravenel's paper proves that the Brown-Comenetz dual of P(1) = BP/p is a spectrum which is not BP-acyclic, but is K(n)-acyclic for all 0 leq n leq infty

@skd Thanks -- so $\langle MU \rangle \neq \langle \mathbb S \rangle$ because $I_\mathbb Z$ is not $MU$-local, and OTOH $\langle BP \rangle \neq \langle \vee_{n=0}^\infty K(n) \rangle$ because $I_{\mathbb Q / \mathbb Z}(BP/p)$ is $K(n)$ -acyclic but not $BP$-acyclic.