Homotopy Theory

8:07 AM

I did not have time to point it out the day it came out, but this paper by Dylan Wilson and Jeremy Hahn is truly great. I think anyone interested in equivariant homotopy theory should read it (and maybe if you are not interested in equivariant homotopy theory, this will change your mind).

skillpatrol

in CRUDE, 26 mins ago, by skullpetrol

reheating old ideas may lead to an unexpected discovery

Dylan Wilson

3:04 PM

Thank you Denis! That’s very nice of you to say :)

Denis Nardin

4:21 PM

@DylanWilson Thank you (and Jeremy) for writing such an outstanding short paper! We need more short but great proofs like this (and yeah, I know I'm not exactly a shining example here...)

Dylan Wilson

4:35 PM

I’ve always liked reading short papers, and recently have discovered that writing short papers is also more pleasant than writing long papers, so it’s a win win really :)

And thanks again! Also I’m happy you wrote your long papers because, as I’m writing the account of c2 power operations with the proofs, it means I get to liberally cite your thesis instead of adding an extra 20 or so pages ;)

Charles Rezk

5:41 PM

So, if I have a symmetric monoidal $\infty$-category $C$, I can consider its maximal subgroupoid $C^{\mathrm{core}}$, which is an $E_\infty$-space. So I can form its group completion $(C^{\mathrm{core}})_{\mathrm{gp}}$, which is also an $E_\infty$-space. Finally, I can take the pushout of $(C^{\mathrm{core}})_{\mathrm{gp}} \leftarrow C^{\mathrm{core}} \to C$ in symmetric monoidal $\infty$-categories.

Call it $C_s$. Have people thought about this systematically? I would assume so ...

A nice example is $C=$ the Stiefel category. I.e., objects are finite dimensional real inner product spaces, maps are isometric linear maps, monoidal structure is orthogonal direct sum.

Then $C_s$ is a sort of category of "virtual vector spaces and embeddings of them".

The space of maps $m\to n$ in $C_s$ is probably $O/O(n-m)$ for $m\leq n\in \mathbb{Z}$.

Tim Campion

6:54 PM

@CharlesRezk This should be the group completion of $C$ no? I.e. $C_s$ should have the property that every object is invertible under the monoidal structure, and universally so.

Assuming that's correct, I once asked this question relatedly

Charles Rezk

Why so?

Every object is invertible, but not every map is invertible.

Tim Campion

Yeah, I just mean that the objects are invertible, not the maps

Charles Rezk

And it seems there should be a non-trivial monoidal functor $C_s\to \mathrm{Spectra}$, sending $V\mapsto S^V$.

I never heard that called the group completion, but ok.

Tim Campion

yeah, the thing I noticed was some conditions satisfied by lots of examples one might think are interesting which ensure that this thing ends up being trivial.

But the Stiefel category has its morphisms cut down enough that the conditions I worked out don't apply

but i've never seen anyone study this systematically, so i agree there's not really a standard name

Charles Rezk

I see

Tim Campion

7:03 PM

I think there might be a paper by Sagave and Schlichtkrull on a category that sounds like the one you're talking about coming from the Stiefel category

They have several papers together, but I think this might be the one I'm thinking of.

They consider Quillen's $S^{-1} S$ construction in some generality.

And I think that may be an alternate construction to use on the Stiefel category and get the same thing

Charles Rezk

Definition 2.6 in arxiv.org/pdf/1410.4492.pdf ?

Tim Campion

Ah, nice!

Denis Nardin

I kind of would like this to be the ∞-category of separable Hilbert spaces and Fredholm maps, but I have no idea if this is the case

Charles Rezk

Maybe not exactly that.

Denis Nardin

On the other hand it doesn't look like all objects are isomorphic so probably the guess is wrong

On the other other hand I have no idea of what the pushout in symmetric monoidal categories is doing so... who knows?

Hrmm... I guess that the category of separable Hilbert spaces and Fredholm maps is just BO, so that guess is probably wrong

Tim Campion

7:15 PM

By the universal property of the pushout, the image in $C_s$ of any object in $C$ will be invertible.

So if there are non-invertible objects, they would have to not be in the image of $C$

I don't think there are any of those -- if there were, you could just throw them out and get a smaller category which looks like it has the same universal property....

That's why I think that $C_s$ is just the universal thing on $C$ where every object is invertible.

The pushout in symmetric monoidal categories can be computed by a horrible bar construction in $Cat$, so there's that at least....

Charles Rezk

Here's a possible model: objects are pairs $(U,n)$ where $U$ is a universe and $n$ an integer. Morphisms $(U,n)\to (U',n')$ are pairs $(V,f)$, where $V\subseteq U'$ is a subspace of dimension $n'-n$, $f\colon U\to U'$ an isometry such that $U'=V\oplus fU$. Monoidal structure is $(U,n)\oplus (U',n'):= (U\oplus U', n+n')$.

I think in my example, if you model symmetric monoidal $\infty$-categories by $E_\infty$-objects in complete Segal spaces, then when you compute the pushout as one of $E_\infty$-objects in simplicial spaces, it turns out to automatically be a complete Segal space as well.

Tim Campion

This is the same as $(Stiefel^{core})^{-1} Stiefel$ via Quillen's construction (as in Sagave-Schlichtkrull), right? And I think it's not bad to show that Quillen's construction has the right universal property -- the universal thing on $C$ where objects of $C$ become invertible...

(setting aside whether everything is invertible here)

Charles Rezk

I don't know what Quillen's construction is.

Tim Campion

Ah

I actually don't know whether to call it Quillen's or Grayson's

since it appears in Higher Algebraic K-theory II

But it's basically like you said -- if $S$ is a symmetric monoidal groupoid an $X$ has an $S$-action, then $S^{-1} X$ is the universal cateogry-with-$S$-action where the $S$-action becomes invertible

Quillen/Grayson gives an explicit construction for ordinary 1-categories

And presumably Sagave-Schlichtkrull spell it out for enriched categories...

But it's like you said, a category of pairs, where morphism spaces have a twist by the action of $S$.

Charles Rezk

Is that what I said?

Tim Campion

7:28 PM

Yeah, I think so

Well, I suppose it wouldn't explicitly have a universe

Denis Nardin

@TimCampion Are you sure it has the right universal property? What's the reference?

Tim Campion

@DenisNardin I'm actually not sure, but I wrote up some notes at some point to convince myself

The one awkward thing is that Grayson/Quillen quotient their morphism spaces at one point, but it turns out that under the hypotheses they put on the category, they're just taking $\pi_0$ of a contractible groupoid

Denis Nardin

@TimCampion I vaguely remember the S^{-1}S construction having some problems... if feels a lot like the mapping telescope and we know that's not always the right thing

Tim Campion

Yeah, in order for the construction as stated to give the correct universal property, it's important that the technical hypotheses be satisfied -- $S$ should be a category of monomorphisms and translation should be faithful

But basically, if you just don't take a quotient at the point they do, and instead work with a $(2,1)$-category, I convinced myself at some point that you get the correct universal property.

Charles Rezk

Oh I see

Tim Campion

7:33 PM

But when I say "the correct universal property", I mean that you're inverting the action of $S$

There is still work to show that this is the same as making the objects of $S$ invertible when $X = S$

Maybe that's what you're referring to @DenisNardin?

I don't know whether you mean something about the mapping telescope not being the correct homotopy colimit (which sounds like what I'm talking about), or about the "localization as a module" coinciding with "localization as a ring"

Denis Nardin

@TimCampion Let me give you an example: let $X=\mathrm{Sp}^\omega$ the ∞-category of finite spectra and $S$ be the free monoid on one generator acting as multiplication by $\mathbb{S}\oplus\mathbb{S}$. Then $X[S^{-1}]=0$ (both considering the localization as a module and as a ring)

Charles Rezk

If I let $B=C^{\mathrm{core}}$ in what I wrote above, then I can form (in the $\infty$-category sense) what Grayson calls $\langle B,B\rangle$. But apparently it turns out that $\langle B,B\rangle = C$.

Denis Nardin

The mapping telescope however has non-zero K-theory

Charles Rezk

I.e., the morphisms in the Stiefel category are completely determined by the monoidal structure on the core.

Tim Campion

@DenisNardin The mapping telescope being the colimit of $X \to X \to \dots$ where we act by the generator?

Denis Nardin

7:39 PM

Indeed

Essentially, the only additive category where multiplication by 2 acts invertibly is the zero category

Tim Campion

So it's a failure of having the correct kind of cofinality or something?

Denis Nardin

It's a "failure of symmetry". Marco Robalo's thesis has more details, but in short if you write down the "obvious" diagram expressing the invertibility of the map on the mapping telescope, it just doesn't commute

To make it work you need that the cyclic permutation on $n$ elements acts trivially on $L^{⊗n}$ for some $n>1$ (where $L$ is the element you're inverting)

Tim Campion

@CharlesRezk I thought $\langle B, B\rangle$ should always come out trivial?

@DenisNardin I will have to read about this

@CharlesRezk Note that $S^{-1} X = \langle S, S \times X \rangle$

Charles Rezk

Let $B$ be the non-negative integers, viewed as a symmetric monoidal $\infty$-groupoid under addition. Then it looks like $\langle B,B\rangle$ is equivalent to the poset of non-negative integers under addition.

Likewise, with $B=\coprod_n BO(n)$, it seems that $\langle B,B\rangle$ is equivalent to the Stiefel category.

Which means that I can take something like $B=\coprod_n \mathrm{hAut}_*(S^n)$, and get a kind of "Stiefel category of spheres".

Saal Hardali

8:27 PM

Define a simplicial set by setting the $n$-simplices to be the set of cocartesian fibrations over $\Delta^n$ (with small fibers). Is this simplicial set (categorically) equivalent to the quasi-category of quasi-categories? I have a feeling the answer should be no. Otherwise I find it weird it's not mentioned anywhere because it seems like the smallest model I've ever seen for $\infty$-category of $\infty$-categries.

Denis Nardin

@SaalHardali I'm pretty sure the answer is yes, but even proving it's a quasi-category is incredibly hard

Charles Rezk

@SaalHardali This is an interesting question. Cisinksi has a model for the $\infty$-category of $\infty$-groupoids along these lines, using left fibrations instead of cocartesion.

Saal Hardali

@DenisNardin you mean proving the horn filling?

Denis Nardin

@SaalHardali Yes

Charles Rezk

I gather Cisinski has been trying to prove that generalization.

Saal Hardali

8:29 PM

hmmm, why can't you just fill the horns by doing pullbacks?

Ah nevermind, that doesn't work

Yai0Phah

I wonder whether we have fancier ways to explain stacks.math.columbia.edu/tag/07L7

Tom Bachmann

Is this just E(A_*), with A_* viewed as a cosimplicial group under addition?

Yai0Phah

it seems to me that, given a simplicial object $\Delta^{\operatorname{op}}\to\mathcal C$, compose with $\Delta_{[0]/}\to\Delta$, we get an object. The object described there seems to be the Kan extension along this map again to give back a simplicial object.

Saal Hardali

@DenisNardin
So somehow in an ideal parallel universe it's very straightforward to prove that this simplicial set is a quasicategory and then all the straightning unstraightning is demoted to being a specific technique as opposed to a necessary evil?

Denis Nardin

@SaalHardali I believe we live in that parallel universe (well, "very straightforward" might be pushing it, "reasonably natural" is more likely), we just don't know it yet

Charles Rezk

8:34 PM

Math works the same in all universes

Saal Hardali

@DenisNardin I imagine people work differently though...

Anyway it's nice to have some optimism

Do any theorems about infinity ccategories become much simpler assuming this?

Denis Nardin

@SaalHardali Not that I know of

Saal Hardali

I mean like foundational stuff like in HTT

I think the adjunction between infinity-groupoids and quasi-categories becomes very easy.

Maybe it's not so difficult to begin with though.

I mean the potential is there. With the models in HTT the quasi-categories are all homotopy coherent nerves of simplicial categories. So it becomes very difficult to prove stuff using simplices.

This model is tiny in comparison to all of them.

Charles Rezk

It's not the size of the model that's important I think. It's the ease of carrying out the straightening/unstraightening correspondence.

Saal Hardali

Yeah, I guess you're right. Having it be a taotology cuts the length of a lot of proofs I would imagine.

Charles Rezk

8:43 PM

It's convenient if every cocartesian fibration is a pullback of the universal one up to isomorphism, rather than up to some sort of equivalence, which is all we have now.

This is how the universal principal bundles work: over a nice (i.e., paracompact etc) space $X$, every principal $G$-bundle over $X$ is isomorphic to one obtained as a pullback of some map $X\to BG$.

"Isomorphic" meaning isomorphic in a 1-category.

Tom Bachmann

Can you give an example of how this is useful?

Charles Rezk

It's useful cause then I could just constantly tell myself "it works exactly like principal bundles", so there would be one less thing I have to remember.

Saal Hardali

Isn't it also nice that it makes the mapping spaces between quasi-category exactly those coming from the natural enrichment. I mean $Fun(X,Y)_n = Hom(X \times \Delta^n, Y)$.

Tom Bachmann

@SaalHardali Isn't that already true?

@CharlesRezk Fair enough.

Saal Hardali

Upto equivalence yes. But the actual mapping space between quasi-categories is much bigger.

You take the homotopy coherent nerve of the simplicial category.

Then you take the mapping space.

Charles Rezk

8:58 PM

I would have taken $Fun(X,Y)$ as the definition of the mapping space.

Tom Bachmann

Well perhaps this is exposing my ignorance of foundations. But isn't the Joyal model structure simplicially enriched by your formula? And fibrant objects are quasi-cats, everything is cofibrant, so this is already homotopically correct...?

Alexander Campbell

@SaalHardali That this simplicial set is a quasi-category is proved in Hoang Kim Nguyen's recent PhD thesis (see Theorem 3.4.7): epub.uni-regensburg.de/38448

Saal Hardali

Oh, you are both correct. I got confused about the coherent nerve. I thought it didn't preserve limits for some reasons than I remembered it is a right adjoint rather than a left. So it is the correct formula

I guess what it does help with is in relative situations. If you want to do mapping spaces in functor categories. Then it would matter.

Like mapping spaces in the infinity category $Fun(X,Cat_{\infty})$

Now these are easier.

Charles Rezk

@AlexanderCampbell There is a lot happening in that thesis

Saal Hardali

@AlexanderCampbell Awesome!

William Balderrama

9:11 PM

@AlexanderCampbell That thesis shows the simplicial set is a quasicategory, but is it known whether it's a "correct" quasicategory? (I.e. equivalent to some other quasicategory of quasicategories).

Alexander Campbell

@WilliamBalderrama I believe that is still work in progress.

William Balderrama

I see, thanks.

Charles Rezk

@TomBachmann I think Lurie's setup is like this. He has a particular model for $Cat_\infty$, and a cocartesian fibration $U\to Cat_\infty$ over it. So for any $f\colon C\to Cat_\infty$ you can pullback to get a cocartesian fibration over $C$. This turns out to be literally the unstraightening construction he describes. ...

... but going the other way is harder. You start with a ccocartesian fibration $p\colon E\to C$. Then yuo apply straightening to it to get a simplicial functor $\mathfrak{C}C\to $ (simplicially enriched categories). But the simplicially enriched categories don't come to you as fibrant, so you have to apply some fibrant replacement before adjointing back to get a map $C\to Cat_\infty$.

In some sense, you don't care, as long as you know $p\colon E\to C$ is classified by some $f\colon C\to Cat_\infty$. Except that you will want to have a datum that says "$p$ is classified by $f$", and this datum is hard to get your hands on I think.

In this new model, I guess you have a universal cocartesian fibration $q\colon U'\to Cat_\infty'$, so that the datum of "$p$ is classified by $f$" is just a pullback square $p\Rightarrow q$ in simplicial ses.

Tom Bachmann

I see. Nice!

Saal Hardali

9:26 PM

Lurie's straightning unstraightning gives functors in both directions right?

(I mean from the small model to the coherent nerve of marked simplicial sets and in the other direction)

Charles Rezk

Functors between what?

Saal Hardali

The quasicategory we have now discovered existed and the coherent nerve of fibrant marked simplicial sets.

Charles Rezk

I would think so. If they both classify the same thing "up to equivalence", they should be equivalent.

Saal Hardali

One functor is just the straightning. The other map is the classifying map of the unstraightning functor from simplicial sets to itself.

More accurately it is the classifying map of the unstraightning of the functor from marked simplicial sets to itself given by unstraightning constructions.

Unstraightning over a point that is...

Charles Rezk

There's more that you want. If $p\colon E\to C$ and $p'\colon E'\to C$ are cocartensian fibrations, with a map $q\colon E\to E'$ compatible with them (I think $q$ needs to preserve cocartesian edges?), then there should be a natural transformation $f\to f'$ of functors $C\to Cat_\infty$ corresponding to $q$.

I don't know if that's a consequence of this thing or not.

Saal Hardali

9:34 PM

I actually think it's simpler than what I said.

Let's call our new small model of infinity categories $\mathcal{B}$ (for best).

Then straightning gives a simplicial functor $\mathfrak{C}[\mathcal{B}] \to Set^{+}_{\Delta}$.

Lurie gives us a striaghtning unstraightning adjunction:

$(Set^{+}_{\Delta})_{/ \mathcal{B}} \to (Set^{+}_{\Delta})^{sSet^{+}_{\Delta}}$

If this is an equivalence of simplicial categories, would that be enough?

(after taking fibrant objects in both sides of course)

Just model categories actually.

Charles Rezk

Does the functor take fibrant objects to fibrant objects?

Saal Hardali

Straightning does.

I think...

Charles Rezk

Really?

It's a badass colimity thing.

Saal Hardali

Or do I have it the other way round

Oh yes you're right

Sorry.

Charles Rezk

Unstraightening is the nice one.

Saal Hardali

9:44 PM

Yeah I agree.

Charles Rezk

It's still amazing to me that Jacob never bothers to write down what unstraightening is explicitly.

Saal Hardali

He writes the cosimplicial thingy

That's something

Charles Rezk

What's that?

Saal Hardali

The unstraightning for the case when the category and the simplicial set are both points

It's a functor from simplicial sets to itself.

marked*

Charles Rezk

Oh ok.

Saal Hardali

9:48 PM

Anyway we can take the unstraightning of the identity to get a functor $N[Set^{+}_{\Delta}] \to \mathcal{B}$

which is a cartesian fibration because unstraightning is the nice one

So it is classified by a map from $\mathcal{B}$ to itself.

So the question is whether this map is homotopic to the identity.

I think that's about right...

To sum up:
1. Apply Straightning to the universal cocartesian fibration over $\mathcal{B}$ to get a simplicial functor $\mathfrak{C}[\mathcal{B}] \to Set^{+}_{\Delta}$
2. Use the simplicial functor from (1) to perform unstraightning on the identity from marked simplicial sets to itself to get a cartesian fibration $N[Set^{+}_{\Delta}] \to \mathcal{B}$
3. The cartesian fibration from (2) is classified by a map $\mathcal{B} \to \mathcal{B}$.

Question: is the map from (3) homotopic to the identity?

When I look at it now it looks like something lurie proves.

(in chapter 3)

Actually maybe not...

I think the map (3) sends an n-simplex, cartesian fibration over $\Delta^n$ to the unstraightning of its straightning. I'm not an expert on what's in HTT chapter 3. But I would imagine this kind of thing being relatively well understood given what's there.

In particular if this map from (3) is indeed homotopic to the identity I wouldn't be surprised if the the proof is hidden there somewhere.

Anyway, I'll go to sleep now before I embarrass myself any further.

(when I wrote N[Set^{+}_{\Delta}] I was tripping. It's not literally the nerve of marked simplicial sets. It's just the unstraightning of the straightning of the universal cocartesian fibration.)

Yai0Phah

11:47 PM

@TomBachmann I believe that it is even simpler. Basically speaking, if $X_\bullet$ is a simplicial object in a pointed category, we can associate each $X_n\to*$ a Čech complex, and we get a bisimplicial object. Then we take the diagonal. It is contractible because each Čech complex is. But I fail to figure out what kind of canonical contractible space it is.

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