wow dl.dropboxusercontent.com/u/1741495/papers/tornado1.pdf

Given a map $S^{k}\to GL_1(R)$, for $R$ a ring spectrum, there are two ways I can think of for getting a map $\Sigma^kR\to R$. One: take the map $S^k\to GL_1(R)$ and smash with $R$, and then use the action of $GL_1(R)$ on $R$: $\Sigma^kR\to GL_1(R)\wedge R\to R$. The other (for $k>0$): use the fact that $\pi_k(GL_1(R))\cong \pi_k(R)$ and so smash $S^k\to R$ with $R$, getting $\Sigma^kR\to R\wedge R\to R$, using the ring map. Are these obviously equivalent?

This question arose for me from Tobi and @OmarAntolín-Camarena's paper.

Hey Jon, since the action of GL_1 R on R is by multiplication, these are equivalent.

But $R$ is a spectrum and $GL_1(R)$ is a space?

I mean, I guess I believe that these are equivalent.

But the technicalities escape me. Especially if $R$ is not $E_\infty$.

I actually do not think so. In the second action you are "shifting" $\pi_k(GL_1(R))$ by the multiplication by some unit. I do not know how to make this precise

Ah here it is: take the zero element in $\pi_k(GL_1(R))$. In the first action it acts as the identity, in the second as the zero map

Right... it's somehow the difference between the additive and multiplicative structure on $GL_1(R)$ or something...

I mean, the second is not even an action. If $R$ is an ordinary ring you are sending $r\in R^\times$ to the multiplication by $r-1$

I.e. even if $R$ is $E_\infty$, $\pi_*(gl_1(R))$ is not isomorphic as a ring to $\pi_*(R)$.

$\pi_*(gl_1(R))$ is not even a ring in any natural way

Ah sure. Since its multiplicative structure gets eaten up in turning it into a spectrum.

Hm. So I wonder what precisely Tobi and Omar meant. They go from the data I describe to a map $\Sigma^kR\to R$.

Probably the first, it is the canonical one after all...

The second is not canonical?

One has to choose an iso of homotopy groups?

@CraigWesterland your thoughts? are we going astray here?

The isomorphism of homotopy groups is canonical but it is not induced by a canonical map of the corresponding spaces. All choices of maps are homotopic but not uniquely so. I dunno, it just gives me a bad feeling

Hm. Yeah, I don't like bad feelings.

Sorry, I'd not thought this through very carefully. It kind of depends upon what you mean in your construction \Sigma^kR --> GL_1(R) \wedge R --> R. Really, the second map should be \Sigma^\infty GL_1(R)_+ \wedge R --> R: the action of GL_1(R) translates into an action of the suspension spectrum (with a disjoint basepoint) on R

Right. I definitely should have said that.

But then you have to ask how you're getting the map from S^k into \Sigma^\infty GL_1(R)_+; you have to shift the basepoint

your map was S^k --> GL_1(R) on the space level

Oh. Hmmm.

call that f

then the target of f is based at the unit of R

yeah.

so I can get a new map f-1: S^k --> Q(GL_1 R_+) which is correctly based at the new disjoint basepoint by formally subtracting 1.

I tihnk

the adjoint is a map S^k --> \Sigma^\infty GL_1 R_+

when you multiply this with something in R, that amounts to multiplication by f-1

I see. Where by $S^k$ on the second line we now mean $\Sigma^k\mathbb{S}$

Yep

Better: there is even a map \Sigma^\infty GL_1 R_+ --> R which is adjoint to the identity map on R

in some sense, you're asking how does this behave in homotopy

Which adjunction there?

ABGHR/May

that is:

the \Sigma^\infty,\Omega^\infty one?

if X is a spectrum and R an E_\infty ring spectrum, Map(X, gl_1R) = Map_{E_\infty}(\Sigma^\infty X_+, R)

O I see.

Ugh, not quite

well that would be saying something about being adjoint to the id map on GL_1(R)

not R

Yep

sorry

O ok.

Now, the unstable homotopy of GL_1 R is isomorphic to the homotopy of R in positive degrees. This is the identification that makes Denis uncomfortable

Right.

on the space level, we have GL_1(R) --> Q(GL_1(R)_+) which is the difference (using the infinite loop structure on the target) between the usual inclusion GL_1(R) --> Q(GL_1(R)) --> Q(GL_1(R)_+) (which is not based) and GL_1(R) --> Q(GL_1(R)_+) which is constant at the unit of GL_1(R) (which is also not based); the whole thing is then based

This gives us a map from pi_k(GL_1(R)) --> pi_k Q(GL_1(R)_+)

which I can then compose into \pi_k(R) via the map above

Now I've managed to confuse myself as to whether or not I believe this is an isomorphism in positive degrees.

Haha. Sorry I was away for a sec writing notes about standard deviation for my class.

Don't worry about it Craig. I suspect I just need to go back and spend more time with ABGHR.

@DenisNardin I think this operadic kan extension stuff isn't so bad, at least if you don't try to go thru all the proofs. this formula in 3.1.2.2 seems to basically just be saying you can compute it on objects by taking the colimit over the fiber (in the case that you're extending along a morphism)

and so, if one trusts all the other stuff written down, then one gets a way of constructing left kan extensions that preserve operadic structure, and actually checking that the object is the right one

If I have a simplicial presheaf on a site C, does taking associated sheaves commute with taking pi_0 objectwise?

generally not - pi_0 doesn't commute with limits

isn't sheafification a left adjoint?

and the "pi_0" functor sSet --> Set is a/k/a "colimit"

(thinking of sSet as a 1-category)

yes, but we're applying pi_0 objectwise, so I don't see how it follows from this

@SaulGlasman why is commutativity with limits relevant?

because taking associated sheaves involves taking a limit objectwise

you've got a diagram in Set-valued presheaves (indexed by Delta^{op}), and you're asking about commuting "colimit" with the left adjoint "sheafification"

left adjoint localizations like sheafification often involve taking limits objectwise

and they seldom commute with corepresentable functors

@AaronMazel-Gee ah right, of course - I was thinking of them as presheaves of simplicial sets, but thinking of it this way it's obvious. thanks!

i'm mystified by saul's observation though, the values are indeed defined as limits aren't they

yeah, so here's a silly example

i guess i really just don't understand sheaves

what a wacky world, a left adjoint defined pointwise by taking some limits

take the site which is just a commutative square of objects

X -> Y

| |

V V

Z -> W

i'm still confused about where limits are coming in. sheafification can be described by applying the plus construction twice (or infinitely many times for presheaves of oo-groupoids). and the plus construction is an objectwise colimit...

where Y and Z are designated a cover of W

then the operation of sheafification on a presheaf F is simply replacing F(W) with F(Y) x_F(X) F(Z)

and you can easily cook up examples where this pullback doesn't commute with pi_0

maybe I need to enlarge this category to make it strictly a site but something like this will work

what's the plus construction in this context?

@AaronMazel-Gee in a presentable situation, all left adjoints can be defined pointwise by taking some limits - that's the content of the adjoint functor theorem!

ncatlab.org/nlab/show/plus+construction+on+presheaves

apparently there is also this formula even, which is also a colimit

the plus construction involves limits and colimits

anyway, here's an example of where \pi_0 and sheafification don't commute: take a scheme, take the sheaf of units, deloop it componentwise to get a presheaf of 1-types – \pi_0 of the presheaf is the terminal presheaf, which is already a sheaf, but \pi_0 of the sheafification is the presheaf of Picard groups

although, if you sheafify that you get the terminal presheaf again, hmmm...

@AaronMazel-Gee hm. that seems reasonable. let me think for a bit about it.

yeah that must be true.

because it's saying that we want to take $b$ to something built (more or less) as a colimit over everything that maps to $b$.

in other words, it should be a formula for constructing $\overline{F}\vert B^\otimes(b)$ and if this element is to be dependent on "things living over $b$" then whatever it takes $b$ to must be the colimit... I think I'm talking in circles here... but that seems like the right interpretation Aaron.

@ZhenLin hmm, right. if I ask the question for "homotopical" sheaves of homotopy types, then it's still reasonable, right? as then the sheafification wouldn't just be object-wise

I do mean homotopical sheafification

even then pi_0 of the sheafification will be Pic?

componentwise \pi_0 is the presheaf of Picard groups, "internal" \pi_0 is terminal

ah, so it's ok then

this is a nice bunch of theorems from a group of talbot 2013 participants arxiv.org/pdf/1511.03526.pdf

Lordy. For an $n$-fold loop space $X$, is the associated $\infty$-operad $X^\otimes$ a sifted simplicial set...?

I'm having a hard time seeing how the diagonal morphism of anything can be cofinal...

(which is equivalent to being sifted, apparently)

I feel like this is the kind of thing @ClarkBarwick would know... =/

what's the associated oo-operad you're talking about?

@SaulGlasman well, it's an E_n-monoidal quasicategory right?

In other words, there's a map $E_n\to qCat$ picking out $X^\otimes$

which is a simplicial set

Or... no i guess it's a diagram of simplicial sets.

So then I guess I mean the associated morphism $X^\otimes\to E_n$

ok, I see

hmm, I have no idea whether such a thing would be sifted

Yeah me neither.

why do you ask?

So, a lax $E_n$-monoidal map $BG\to Spectra$ picks out a $G$-action on an $E_n$-algebra, by $E_n$-algebra maps. I want to know if I can compute the colimit of this diagram in the underlying category of spectra.

But... okay maybe I'm mixing things up here.

That doesn't really bring $BG^\otimes$ in at all.

Yeah, hrm. I'm not seeing now why I was worried about that.

So, let's say I have a lax $E_n$ map $BG\to Spectra$ picking out a $G$-action on an $E_n$-algebra $X$. Then I want to know whether or not this functor actually factors through $BGL_1(X)$, I guess...

which would require the maps going to $X$-module morphisms.

We know that they go to $S$-algebra morphisms.

Which means they must be compatible with $X$'s action on itself.

So they must be $X$-module morphisms w/r/t $X$'s action on itself.

So this functor can be lifted to $BGL_1(X)$ and then, by inclusion, to $LMod^{E_n}_X$.

The question I was bothering with was whether or not the colimit computed in $LMod_X^{E_n}$ was the same as the colimit computed in $Spectra$

And this seems to depend on whether or not $BG$ is sifted, but NOT on whether or not $BG^\otimes$ is sifted.

@SaulGlasman when doing operadic stuff, it may also be useful to know that the category $Disk(n)$ (of Ayala, Francis etc.) is sifted. I don't know if this means that something in the shape of $Disk(n)$ (e.g. an $E_n$-algebra) is sifted... but it seems possible?

But I don't think $BG$ is sifted...

BG is definitely not sifted, since sifted $\infty$-categories are weakly contractible

@Jon: the forgetful functor from E_n-algebras in spectra to spectra is a right adjoint, so you expect it to preserve homotopy limits, not homotopy colimits, and in fact it does not preserve homotopy quotients. the homotopy quotient by the trivial action of a group G in E_n-algebras is tensoring with the "E_n group algebra" of G, but the homotopy quotient by the trivial action of a group G in spectra is tensoring with the suspension spectrum of G, which is only the E_1 group algebra

this is all assuming i've correctly understood your question

@QiaochuYuan yeah i think that's definitely right.

Wait. I mean... wait a sec...

Something seems funny about that.

Suppose I have an $E_n$-map $BG\to Spectra$

Might as well pick out $S$ with a $G$-action

there's a map $\ast\to BG\to Spectra$

this always induces (by Thomification) a map of $E_n$-rings $S\to S/G$

where $S$ is $S$ tensored with the suspension spectrum of $\ast$, as you say.

Hrm.... okay maybe I see where I'm going wrong here.

I certainly don't expect that map to preserve colimits, as you say, since it's a right adjoint... I was hoping for some kind of siftedness to get past that... but clearly we don't have that.

really the question is about whether or not it can be lifted to the category of $X$-modules, not $S$-algebras, anyway.

I don't know what X is, but I believe for any R the forgetful functor from R-modules to spectra does preserve both limits and colimits

$X$ is just some $E_n$-algebra in spectra.

We just have some lax $E_n$-monoidal map $BG\to Spectra$ picking out a $G$-action on some $E_n$-algebra $X$ and I think that it factors through $Mod_X$

And moreover, the colimit computed in $Mod_X$ is the same as the colimit computed in $Spectra$

So what's the question?

Whether or not that's true, haha.

It seems like it must be.

That factorization.

B/c of what I said earlier, about the maps all preserving algebra structure, hence preserving $X$'s action on itself, so they're maps of $X$-modules.

@Jon: okay, I don't understand your question then. are you trying to compute a colimit in E_n-algebras or in module spectra over an E_n-algebra?

@QiaochuYuan well, no, your response to my original question was exactly what I was wondering about, but as usual, I was asking my ultimate question in a very roundabout way, sweeping up odds and ends as I went along.

The question, precisely stated, is the following: Given a lax $E_n$-monoidal map $BG\to Spectra$ with $\ast\mapsto R$, is it true that it factors as a lax $E_n$-monoidal map $BG\to Mod_R^{E_n}\to Spectra$, with the second map being the forgetful map? And, morevoer, is the colimit of $BG\to Mod_R^{E_n}$ equivalent to the colimit of the original map $BG\to Spectra$ after forgetting down to $Spectra$?

So is G supposed to be an E_{n+1}-space then?

Yeah.

It's getting pretty late here, but I think 3.4.4.6 in Higher Algebra answers your last question in the affirmative

Agreed.

Why is it clear that R is an E_n-algebra?

Never mind, that's just because you have a unit in BG

Ah. Hm... I guess I was (incorrectly?) thinking that the unique object in $BG$ was an $E_n$-algebra in $BG$ as a quasicategory.

I mean, if $C$ is an $E_n$-monoidal quasicategory then surely the multiplicative unit is an $E_n$-algebra, which is maybe what you're saying.

Yeah, I think that's the same thing as I said - except to me it's weird to say "unique" when the space of choices is not contractible

Ah. I mean, what do you mean by that? Up to homotopy, any model of $BG$ should have one object right?

Hmm, it sounds like you're asking something pretty close to: if A is an O-algebra in C, does this lift to an O-algebra in A-modules in C...

But the choice of that one object is not contractible - it's precisely BG, so it's merely connected

That's basically what I'm saying. But... it's a little tricky here. For instance, depending on how we work things, the category of $A$ modules is at best an $E_{n-1}$-monoidal category.

So something funny is going on here...

Hmm, HA Corollary 3.4.1.7 seems to give what you want, assuming the operad corresponding to the E_n-space BG is still coherent...

No, I think operadic O-modules are still O-monoidal

E.g. for E_1 the operadic modules are bimodules

Well right. I was just kind of realizing that I think I need this result for left modules (not bimodules). Which confuses me about how I'm even getting to where I am.

Hmm, I have no idea what the analogue of left modules is for E_n-algebras

Though I think Geoffroy Horel has a paper where he describes different kinds of modules for operads algebras

IIRC, Lurie shows that if you take an $E_n$-ring $R$, consider the category $LMod_R$, then it admits an $E_{n-1}$-monoidal structure.

Right

And that's what I'm referring to.

In my particular example, basically I've got a lax $E_1$-map $BG\to Spectra$, and I want to apply a theorem that works for maps $BG\to BGL_1(R)$, and I want to show that it's of that form.

But perhaps this simply isn't true in general.

But if it's an E_1-map, then R-modules doesn't have any monoidal structure, so how could the map factor through R-modules?

Yeah, that's what I just realized is the problem.

Ugh what a hassle.

I wonder... maybe there's an associated (non-monoidal) map $BG\to Mod_R$ giving the $G$-action on $R$ as an $R$-module, and the colimit in $Mod_R$ is equivalent to the colimit in $Spectra$ as a spectrum.

Well, the colimit in Mod_R definitely maps to the colimit spectra, at least...

Oh I see... I have an $E_1$-map $BG\to Spectra$ which takes $\ast\to R$, but... $R$ is actually $E_2$

So it's at least possible for there to be an $E_1$-map $BG\to BGL_1(R)$.

So does your map from BG actually land in E_1-algebras in spectra?

Yes.

Well... I mean it must right?

If it's a lax $E_1$-monoidal functor?

No, I mean do you have a map from the operad over E_1 coming from BG to E_1-algebras in spectra?

So if you restrict along the map from a point to BG you get the E_2-algebra structure

Hm. That's not so clear to me. I mean, all that I know that I have is a functor of $E_1$-monoidal quasicategories $BG\to Spectra$

Whose value happens to be E_2?

Yeah.

If the map doesn't respect the E_2-structure, then I don't see why this should be true

Yeah.... I see what you're saying.

Sorry, I know you said it's late there, haha.

Yeah, I think it's time for bed - good luck!