Homotopy Theory

4:03 AM

Every loop space is rectifiable to a strict topological monoid. Does this go further to being a strict topological group? I seem to recall this being true.

Rune Haugseng

4:59 AM

@JonathanBeardsley Yes, I think you can take the realization of Kan's loop group in simplicial sets

Jonathan Beardsley

Ah right, okay. Yeah, you might need to like... deloop it, replace with a reduced space, then Kan loop it, or something, but yeah. Great.

Jonathan Beardsley

5:47 AM

Hm, sort of related question: what is the Dwyer-Kan localization of the model category sSet_0 of reduced simplicial sets (whose w.e.'s and cofibrations are inherited from sSet)? I guess it's "pointed ∞-groupoids," but it feels like it could also be equivalent to connected ∞-groupoids?

Ah but maybe the issue is that the space of ways we can homotope a connected ∞-groupoid to one coming from a reduced simplicial set is not contractible.

So we get a fibration $DK(sSet_0)\to Gpd^{\geq 1}_{\infty}$ whose fibers are connected, but not contractible.

Or something like that...

Saal Hardali

9:11 AM

@JonathanBeardsley This is related to something ive been wondering about (since im a noob at simplicial homotopy). What does the free simplicial group on a reduced simplicial set model? Is it this the loop space?

Anonymous

11:41 AM

@SaalHardali: the coherent cohomological dimension of the punctured spectrum $U$ is $n-1$ since $H^{n-1}(U, O_U)$ is nonzero (for example, if $n >= 1$, then by Matlis duality, this group is dual to the dualizing sheaf of $R$, which is always nonzero for a nonzero ring $R$). If $U$ was covered by $<= n-1$ affine opens, then its coherent cohomological dimension would be $<= n-2$, so you get a contradiction.

Saal Hardali

1:09 PM

@Anonymous What do you mean by Matlis duality and (also why does $R$ have a dualizing sheaf? it could be non-gorenstein right?). Anyways I think you are right that already $H^{n-1}(U, \mathcal{O}_U) \cong R^n \Gamma_{ \mathfrak{m} }(R)$ which must be non-trivial for $n = dim R$ for noetherian rings, I don't remember exactly how the proof of this goes to know whether this argument is circular or not.

All in all its comforting to know that "affine covering dimension of the punctured spectrum" is a valid definition of Krull dimension in a noetherian situtation.

Saal Hardali

1:21 PM

Here's an entirely unrelated question:
Given a stable $\infty$-category with a t-structure $\mathcal{C}$ let $\mathcal{A}$ denote its heart. Then there's a canonical functor $F: D(\mathcal{A}) \to \mathcal{C}$ (maybe not in this generality, though at least I hope so). Let $\mathsf{A}$ be the full subcategory of $\mathcal{C}$ spanned by shifts of the objects in the heart. What exactly should this category satisfy in order for the $F$ to be fully faithful? I think something like "generated by cup products from Ext^1" should be enough but I'm not sure how to make this precise... does anyone kn…

Denis Nardin

2:06 PM

To have that functor you need $\mathcal{C}$ to be presentable, or something like that (certainly at least cocomplete)

Omar Antolín-Camarena

2:41 PM

Does anyone know of a friendly reference for the following fact: if a diagram X(i) indexed by a poset satisfies that colim_{j<i} X(j) --> X(i) is a cofibration for all I, then colim X(i) is also a hocolim? We use it in a section of a paper that ideally should be accessible to combinatorialists, so I'd like a reference that's just for posets and not, say, Reedy categories.

Rune Haugseng

3:06 PM

@JonathanBeardsley Hmm, I think I answered a different question than the one you asked...

Drew

3:22 PM

@SaalHardali Under some conditions (left complete t-structure, and $\mathcal{A}$ having enough projectives) you get a functor from the left bounded derived category of $\mathcal{A}$ to $\mathcal{C}$. In Proposition 1.3.3.7. of HA Lurie gives conditions to ensure that this functor is fully faithful. Maybe you can get something out of this.

ufabao

3:57 PM

Does anyone know how to do the part of Corollary 2.18 in arxiv.org/pdf/1703.07842.pdf that's left to the reader?

Tim Campion

6:43 PM

So in unstable $K(n)$-local homotopy theory there's this category you get by looking at $d_n$-connected pointed spaces and localizing them at the Bousfield-Kuhn functor. According to Heuts, the resulting category is precisely Lie algebras over the $K(n)$-local stable category. Now, $d_n$ can be anything, as long as it's "sufficiently large", but I'm wondering: what's the minimum you can take it to be? Maybe $n+1$?

The reason I would guess $n+1$ is that if you take pointed spaces and localize at $K(n)$-equivalences, it appears that $n+1$-connectivity is the threshhold where a space starts to depend only on its connective covers (since $\tilde K(n)_\ast K(\mathbb Z, m) = 0$ for $m > n+1$). But I'm hazy on the relationship between $K(n)$-localization and Bousfield-Kuhn localization.

Saal Hardali

6:59 PM

@TimCampion I'm surprised that simply connected enough for this, why doesn't d_n=1 work?

Tim Campion

@SaalHardali I'm a bit hazy on the details, but it has something to do with the construction of the Bousfield-Kuhn functor and the connectivity of the type $n$ spaces you use, I think.

Saal Hardali

Ah wait, you want this localization to kill low eielenberg maclane spaces right?

Tim Campion

I think so

Saal Hardali

otherwise you get a bigger category probably

Tim Campion

I just don't know what the threshhold could possibly be if it doesn't just come down to killing the EM spaces, basically.

Saal Hardali

7:03 PM

Is there a diffeeent notivation for taking the connective cover other than making this theorem true?

Tim Campion

It's a move that already appears in the papers of Bousfield and Kuhn where they define the Bousfield-Kuhn functor.

I guess maybe I should look there to see why they do it.

Saal Hardali

I should check this paper at some point too

I thought bousfield kuhn functor was defined regardless of connective covers.

That the loop functor had a left inverse.

(out of T(n)-local).

Tim Campion

I think the way it works is that the Bousfield-Kuhn functor, when defined on all spaces, simply factors through the $d_n$-connective cover. (Of course, everything here works for either the telescopic or the $K(n)$-local case.)

Anonymous

7:47 PM

@SaalHardali: by dualizing sheaf I mean the lowest nonzero cohomology group of the dualizing complex. This exists as long as the ring is not too pathological. In particular, it exists for complete noetherian local rings. You can apply the reasoning of my previous comment first over the completion, and then pass to the non-complete ring using faithful flatness of completion (you just need that the preimage of the punctured spectrum is the punctured spectrum)

Saal Hardali

8:14 PM

@TimCampion I always thought that it factors through arbitrary connective cover.

The T(n)-localization of the m-connective cover of a T(n)-local spectrum gives back the spectrum for any m.

So at least its reasonable for me that the BK would factor through arbitrary connective cover.

@Anonymous So waht is the version of Matlis duality you are using for the complete local ring? that the local cohomology of R is linearly dual to the dualizing complex?

Jonathan Beardsley

8:50 PM

@RuneHaugseng haha it's entirely possible you were just answering a question i meant to ask, or should have been asking.

Jonathan Beardsley

9:33 PM

@SaalHardali I don't think this is going to be the loop space. There is a Quillen equivalence between reduced simplicial sets and simplicial groups given by Kan loops G(-) and the W-bar construction, W(-). G is the left adjoint and W is the right adjoint. There is also a forgetful functor from simplicial groups into simplicial sets that has a left adjoint, but I don't see why this would be equivalent to G

I can't immediately find a reference for "free simplicial group" but if it means something like "take the free group on each set of simplices" then I think is going to be quite large and not the loop space.

Tim Campion

@SaalHardali @JonathanBeardsley Topologically, the free topological group on a pointed connected space $(X,x)$ is the same as the free topological monoid on $(X,x)$ is the same as $\Omega \Sigma X$. So presumably the free simplicial group on a pointed connected simplicial set $(Y,y)$ is the same as the free simplicial monoid, is $\Omega \Sigma Y$.

Here I mean "free" in the sense of the James construction where the basepoint becomes the unit of the group / monoid

Jonathan Beardsley

Yeah, that seems like it's probably the right analysis.

If we pass to working in the associated quasicategories that's definitely what's going on. I just get nervous about, like, actual simplices.

Tim Campion

I suppose if you want to be careful, what you'd want to check is that the free simplicial group functor is a left Quillen functor $sSet_\ast \to sGrp$. Then because every object of $sSet_\ast$ is cofibrant, it preserves weak equivalences so doesn't need deriving.

Maybe being reduced messes with this though.

Transcript for