@RuneHaugseng shouldn't this only be the case if your automorphism is O-monoidal? for a baby case, consider an Ass-monoidal 0-groupoid, i.e. a set equipped with the structure of a monoid. an automorphism of its underlying set shouldn't necessarily lift to an automorphism of the monoid

(not that "O-monoidal automorphism of the underlying \infty-category C" really has much meaning in the first place)

@AaronMazel-Gee but it seems like if you've got a monoid structure on a set A, and a bijection between that set and another set B, then it induces a monoid structure on B, and that morphism is a monoid isomorphism for the induced structure, yes?

i.e. if the bijection is $f:A\to B$, you define $b\times_B b'=f(f^{-1}(b)\times_A f^{-1}(b'))$ yes?

@JonBeardsley yes, certainly

but i thought the question was about an arbitrary automorphism of i guess B here, and obtaining from it an automorphism of a monoid. but that's not what's happening: you're just renaming elements

err, not the question, but rather rune's assertion that i was disagreeing with

well, right. I mean, in the case Rune and I were discussing. you've got an equivalence of C, and C has an O-monoid structure, and this induces an O-monoid structure on C by way of the automorphism (presumably an equivalent one) and the question is whether or not the automorphism can be lifted to an actual O-monoidal morphism.

(I think?)

what do you mean "you've got an equivalence of C" here?

i mean you've got a cocartesian fibration of infinity operads $X\to Fin_\ast$ (we'll just use Fin_* for now), such that $X_{\langle 1\rangle}\simeq C$, and then I have an equivalence of quasicategories $C\to C$, I guess.

my original question was about an equivalence $C\to D$ and the induced monoidal structure on $D$

okay sure. all i mean to say is this: given two O-monoidal categories C and D, and an equivalence C = D of underlying categories, there's absolutely no reason for this equivalence to lift to an equivalence of O-monoidal categories

so, there are two (equivalent) symmetric monoidal structures on $C$, the one induced by the equivalence $X_{\langle 1\rangle}\simeq C$ and the one induced by the equivalence $X_{\langle 1\rangle}\simeq C\to C$

Oh yes, I think @RuneHaugseng and I both agree with that statement.

But when the O-monoidal structure on the target is the induced one, there should be a way to lift the morphism, do you think?

let's go back to calling the target D. then yes, if you're defining the O-monoidal structure on D to be transported across the equivalence $C \xrightarrow{\sim} D$, then certainly that equivalence is itself canonically O-monoidal. it's when there's an a priori different O-monoidal structure on D where it doesn't come for free

Yep. Then we're in agreement.

But how do you build the structure of the O-monoidal morphism into $C\to D$?

okay, cool

you're only starting with a map of quasicategories.

it's kind of a silly question, I think.

an O-monoidal functor from C to D is just a map of infty-operads over O (maybe preserving inert arrows, whatever) commuting with the equivalences from the level-1-things

well right, so do i use the same infinity operad for C and D?

I'm starting with an infinity operad that has C over <1>, and I know C is equivalent to D

but in your case, the level-1-things are both just $X_1$, so the map on infty-operads is the identity on X and the triangle canonically commutes

right. well, both equivalent to X_1

which i know is, as i said above, a silly distinction to make...

but these sorts of things have been troubling me a bit lately.

and i guess this depends now on whether you're working in the simplicial model category sSet with the Joyal model structure, or whether you're actually working in the quasicategory

(of quasicategories)

let's be point-set

fix an infty-operad O, and let $(X_\bullet \to O, X_1 \xrightarrow{\sim} C)$ and $(Y_\bullet \to O, Y_1 \xrightarrow{\sim} D)$ be O-monoidal infty-categories

yes

then a morphism should be a map $X_\bullet \to Y_\bullet$ over O and a map $C \to D$ so that the evident square commutes

the verticals in the square being the inclusions $C\hookrightarrow X_\bullet$ and the analogous one for $D$?

no, $X_1 \to C \to D$ and $X_1 \to Y_1 \to D$

(that's the whole square)

oh i see.

so we have this map $X_\bullet\to Y_\bullet$ and we restrict it to $X_1$ and $Y_1$. Ok.

so in your case, we're taking $Y_\bullet = X_\bullet$ equipped with $Y_1 = X_1 \to C \to D$

Right.

so the "map over O" is the identity on $X_\bullet$ and "the evident square commutes" because the O-monoidal structure on D was defined so that it does

Yeah. I see that.

Great. Basically just definitions.

yes

in the interest of full disclosure, though, i just made up that definition; i haven't read HA chapter 2 in quite a while. so if you're writing this down (rather than just trying to understand it), you should probably find a reference to make sure it's true. in any case, this assertion will certainly remain definitional.

no no it's fine. i get the basic idea.

i'm not writing it down.

i think it's too basic to really write down anyway.

it's basically checking that "up to homotopy" really is good enough in quasicategories.

and it's kind of a trick, using the identity morphism. you're not, like, building some massive morphism out of this equivalence.

@QiaochuYuan I made a discussion page for us because I would like some help from you.

@JonBeardsley i'd say that it actually reflects the underlying point: that this is "just" the identity map, you're not building any actual automorphisms here

does anyone know of conditions on an algebra object in a monoidal category guaranteeing that all its module actions are via equivalences? for instance, a module M over a field F has that the action map F(x)M --> M is an isomorphism

i'd like it to just be "the algebra is idempotent" (e.g. F(x)F = F), but i don't see how to prove this

...which means that Free is essentially surjective, so actually every module is free, so it works! okay, good :o)

I don't have an answer to your question but can I ask a question? by (x) you mean \otimes, right? also, where is this tensor product taken?

oh, I see you figured it out on your own :)

@pro yeah, i try to avoid using mathjax when it's easy enough to handicraft the desired symbol

but it's taken in the monoidal category C. so in my example, it's tensor product of abelian groups

but then why is F (x) M ---> M an iso? sorry, if it's tensor of abelian groups this can't be right, no?

hmm, i was actually just thinking about F = Q. if M is an abelian group that underlies a Q-vector space, then Q(x)M is just M again, i think since every element of M is already uniquely divisible

so Q (x)_Z Q = Q? really? I didn't know that.

weird

yeah, e.g. you can canonically turn every tensor into one with a 1 in the first slot

cool, makes sense to me, thanks.

there's probably a nifty way to see it algebro-geometrically in terms of QCoh(Spec(Z)), too

these tensor idempotent modules are great because they're the ones that give rise to smashing localizations

how can you see that?

but isn't this discussion special about Z and Q?

the statement is that a module is idempotent if and only if it's the localization of the unit wrt some smashing localization, and I think that's almost tautological

the localization itself, of course, being given by tensoring with the module in question

i still haven't really made my peace with non-smashing localizations. i want to say that this means we have the "wrong" bousfield localization, or even maybe we have the "wrong" definition of E-homology. i don't suppose there's a way of adding a topology or something that refines the notion of "E-homology equivalence"?

@pro well, spectra are pretty close to Z-modules

but actually saul doesn't appear to be specializing to any particular monoidal category, and i guess he needn't

sorry, I meant your original observation of F(x)M being isomorphic to M for any field. I still don't see why that's true.

oh, well if nothing else we can probably work it out elementwise

yeah, so in a tensor a(x)b, if neither is 0 then both are units, and again we can force a=1

it's still true what Aaron said about turning any tensor uniquely into one with 1 in the first slot

beat me to it

to be clear, here M needs to be (the underlying abelian group of) an F-module

@AaronMazel-Gee my feeling is that the non-smashingness of many localizations is a necessary part of the complexity of the world

well, it's purely an issue on non-finite spectra, which makes me want to repair it by saying "ind" somewhere -- that's all

I made a claim about being able to approximate a localization by a smashing localization that way, but actually I don't think it's true

but wait, given an arbitrary smashing localization L : C --> LC, the monoidal product in LC is the completed monoidal product, so we only have $L1 \hat{\otimes} L1 := L(L1 \otimes L1) \simeq L1$ -- that doesn't prove that L1 itself is idempotent

wait, one sec i think i'm being stupid

yeah haha, smashing means $L(-) = L1 \otimes -$

the thing is that once you localize a finite spectrum it almost always ends up non-finite

so you're forced to consider non-finite things

so L1 is at least equivalent to its three-fold monoidal power

can you reduce it to two?

the functor L regarded as a functor C -> C is given by $X \mapsto X \otimes L1$

mm, actually yes: $L1 \otimes L1$ is then local by definition

yeah

haha that almost feels circular

but so then much of the interest of a localization is in what happens to the infinite objects, which is not determined by what happens to finite things at all

think of Q-localization of abelian groups: ok, so you localize Z and you get some weird infinitely-generated abelian group, but the really interesting thing is that if you localize this huge thing then nothing additional happens

this is smashing, but I could equally have said p-completion

i guess it's all bound up in how the monoidal structure does or doesn't commute with colimits

also

oh, no never mind

i was just thinking about whether A(x)- is a left adjoint. so hmm: if the monoidal structure commutes with colimits separately in each variable.....i guess that doesn't guarantee that the right adjoint is an inclusion

another way of thinking about this that I like

most functors that we encounter preserve filtered colimits, and those are nice, but they keep us in the realm of things that are more or less determined by generators and relatons

if we want to transcend that kind of stuff, then we need some non-filtered-colimit-preserving functors, and localizations are some of our best examples of those

@AaronMazel-Gee You're quite right, what I was saying was kind of silly - if you have two different (but equivalent) monoidal structures, then you haven't identified the end points of the monoidal equivalence, so you don't get an action of the automorphisms of C!

I once read that one possible definition of a reduced homology theory is a homotopy invariant functor from pointed spaces to graded abelian groups that maps a cofiber sequence $A\to X\to X/A$ to an exact sequence $H(A)\to H(X)\to H(X/A)$ and such that $H(pt.)=0$.I don't see how this can be right, with this formulation there doesn't seem to be any reason to (naturally) connect $H_{n+1}$ with $H_n$ (in the form either of a boundary homomorphism or of a suspension isomorphism, which is equivalent).

What could that person have meant?

@lenticcatachresis: $X \to CX \to \Sigma X$ should do the trick, I think

When I say an exact sequence $H(A)\to H(X)\to H(X/A)$, I mean levelwise exact sequences $H_n(A)\to H_n(X)\to H_n(X/A)$

So I don't think that helps

you're correct, if you pick one different cohomology A[n] for each n then defining H_n(X) = A[n]_n(X) satisfies that definition. you probably want the suspension isomorphisms H_n(X) ~= H_{n+1}(Sigma X) as part of the data

(if your goal is being somewhat minimalist with the axioms)

Ah, good, that's what I thought, and that's a nice way to see why it's correct. Thanks!

@lenticcatachresis: ah, clearly you did. sorry!

I was thinking this was true, but now I'm questioning myself: a map $X\to BGL_1(R)$, where $BGL_1(R)$ is a subcategory of $LMod_R$, gives me a Thom spectrum by taking the colimit in $LMod_R$. On the other hand, I can take a space $Y\to X$, turn this into a functor $X\to Top$, then stabilize $X\to Spectra$ then tensor with $R$ to get $X\to LMod_R$. Will the latter thing recover everything I can get from the former thing?

I just cannot keep these two effing things straight it seems...

Hey @JustinHilburn I feel like I recognize your name. Did we meet at a WCATS in Eugene a few years ago?

(on oo-categories?)

@JonBeardsley i (still) don't know much about thom spectra, but it looks to me like that only recovers thom spectra of maps $X \to BGL_1(R)$ that factor through the map $BGL_1(S) \to BGL_1(R)$ given by $- \otimes_S R$. since this latter map surely isn't an equivalence in general, i'd conclude that there should be maps $X \to BGL_1(R)$ (and hence thom spectra) that don't arise in this way

yeah.... that seems true to me as well.

in fact, the only way you're getting an R-line (instead of just a parametrized R-module) is if your original map Y --> X is an S^0-bundle, in which case you can only pick out the $\pm 1$ automorphisms of the sphere anyways... so yeah, definitely not

right. yeah because those are the only morphisms you have in BGL_1(S)

hrggg

@JonBeardsley Yes.

@JustinHilburn o cool! what are you up to now?

@JonBeardsley Just about to graduate. I have temporary position at Perimeter for the Spring and Summer and then I need to find something else to do.

ah cool. me too.

@AaronMazel-Gee I think I'm just confusing myself. My brain doesn't work anymore. I guess the real equivalence is this: I have $X\to BGL_1(R)$, which corresponds to $BGL_1(R)\to Top$, which we then stabilize to get $BGL_1(R)\to Spectra$, and ando-blumberg-gepner give a functor $Fun(BGL_1(R),Spectra)\to Mod_R$. That's what I was thinking of. This functor is equivalent to taking the colimit of the original thing $X\to BGL_1(R)\to Mod_R$

that sounds really unlikely to me

the thing you're stabilizing is the straightening of $X \to BGL_1(R)$, so the spectra you get out are the suspension spectra of the fibers of this map

what's the functor $Fun(BGL_1(R),Spectra) \to Mod_R$ supposed to be, anyways?

it's the counit of an adjunction

there's this adjunction that takes (certain types of) categories to their underlying grouplike picard spaces

or, not underlying.

but, for instance, it takes $Mod_R$ to $Pic_R$

or, that's one side of it

okay, so the counit should just be a map $Pic_R \to Mod_R$?

right sorry, i misspoke

not quite

that's one half of the adjunction, what i said

sure, the right adjoint i'd guess

the other half is given by taking a grouplike O-monoidal space to (spectral) presheaves on it

yeah. and this second one is the left adjoint.

surely "grouplike" doesn't make sense for an arbitrary O? only one with a map to Ass, or something

yes

sorry

see theorem 1.6 arxiv.org/pdf/1112.2203v2.pdf

it's pretty crazy, in my opinion. it's a really slick way of constructing thom spectra.

okay, cool

i want to write a chromotopy post called something like "A Tale of Two Thom Spectra"

(1) because i keep getting confused about these two ways of doing it.

and (2) because I think the ABG method is really neat, and kind of revolutionary in a way.

so if you have a map $X \to BGL_1(R)$, i think you just consider this as a point of $Fun(BGL_1(R),Spaces)$ and apply this counit map -- no stabilization anywhere

I think the stabilization is Corollary 8.1

i'm deeply confused then, i see no reason why the colimit of the diagram $X \to BGL_1(R)$ should factor through the fiberwise stabilization of this map (thought of as a parametrized space)

rather, the construction of that colimit

so, if i understand correctly, they're basically reusing the argument from thm 7.7 but replacing Pr_L with stable presentable categories (and colimit preserving morphisms) and the category of spaces with the category of spectra, I think.

of course that proof, thm 7.7, isn't really a proof.

intuitively it doesn't seem that bad to me though. note that thom spectra are always connective. in fact they are basically twisted suspension spectra. so it's not so hard to believe that they factor in this way, at least to me.

that seems pretty orthogonal to questions about stabilizing the source diagram-shape

okay, so a baby case where i believe the statement is when $X \to BGL_1(R)$ is constant at the basepoint. then the thom spectrum should be the free R-module $X \otimes R$. this clearly factors through the suspension spectrum functor (in the X-variable)

i guess maybe this should be the intuition for the more general case