Homotopy Theory

Omar Antolín-Camarena

3:41 AM

@TimCampion I like it! But even so it would take some getting used to.

Harry Gindi

5:43 AM

@JonathanBeardsley What if you restate the ascending chain condition in terms of chains of surjections?

but I dunno

Girish

10:31 AM

I was wondering: if $Y$ is a fibrant simplicial set and $\Delta^{\bullet}$ is the cosimplicial simplicial set, is $Y(\Delta^{\bullet})$ a (Reedy) fibrant bisimplicial set?

Saal Hardali

5:12 PM

I have a weird question:

Lets take the directed diagram whose objects are derived categories of sheaves of vector spaces (not necessarily locally constant) on a ball and functor between them corresponding to restrictions along inclusions between them (i.e. i'm taking the colimit as the radius goes to zero).

What is this category?

All the categories in the diagram are isomorphic so it's a bit like defining spectra from spaces (only here wer'e taking a colimit and not a limit). In slightly less precise term this is the stalk of the sheaf of categories of sheaves at a point on any manifold.

Maybe not sheaves on the unit sphere but rather $\mathbb{R}_{\ge 0}$ equivariant sheaves on $\mathbb{R}^n$.

(worth mentioning i'm taking the colimit over shrinking open balls)

*$\mathbb{R}_{>0}$-equivariant

Saal Hardali

5:42 PM

Decided this was worth to ask on the main site...

1

Q: The stalk of the sheaf of sheaves of vector spaces

For $X$ a reasonable topological space denote by $Sh(X)$ the derived (stable $\infty$-) category of sheaves of vector spaces (over a fixed field $k$) on $X$. Consider the filtered diagram whose whose objects are $Sh(U)$ for all $0 \in U \subset \mathbb{R}^n$ open subsets. And morphisms are restr...

Charles Rezk

6:47 PM

@SaalHardali I don't know the answer. But why in particular "derived sheaves of vector spaces"? (Do you mean sheaves of complexes of vector spaces?) What if $\mathrm{Sh}(X)=$ sheaves of sets on $X$. Do we know the answer in that case, e.g., the stalk of $\mathrm{Sh}(\mathbb{R})$ at a point?

Charles Rezk

6:59 PM

And I assume you mean colimit in the $\infty$-category of $\infty$-categories?

Denis Nardin

If you want to be like "spectra" from "spaces" you need to take the colimit in the ∞-cat of presentable ∞-cats

Saal Hardali

Yes, I mean colimit of infinity categories and Yes, i mean complexes of vector spaces (just the classical derived category considered as an $\infty$-category). I chose vector spaces for no reason other than assuming the question would be easier to answer in this form and the fact that I cape up with it by thinking about sheaves of vector spaces so I didn't generalize beyond this context.

Perhaps I want to take colimit in presentable categories i'm actually not sure about that point.

One more reason I phrased it for vector spaces is that I thought initially that this question could be approached with microlocal sheaf theory.

In any case the answer for sheaves of spaces would give an answer for sheaves of vector spaces if we take the colimit in presentable categories.

Since tensor commutes with colimits and Sheaves of vector spaces is the tensor product of sheaves of spaces with vector spaces

Maybe that's a compelling reason to take the colimit in presentable categories

Saal Hardali

7:24 PM

I deleted the question since I realized I need to think about this more carefully,

Thanks to a comment by Will Sawin

Tim Campion

@SaalHardali It should be pretty easy to disambiguate whether you want the colimit in categories or presentable categories, as they are wildly different. Colimits in $Cat$ start by taking a big disjoint union of cateogories and building from there, whereas colimits in $Pres^L$ are computed by taking right adjoints to everything and then taking the limit in categories.

Saal Hardali

Wait, if what you say is true then It looks like the colimit in presentable categories is jmuch smaller and my initial guess seems a lot more reasonable now

Denis Nardin

I think that the colimit in presentable ∞-cats is the category of sheaves on R^n supported at 0 (i.e. such that the restriction to R^n\0 is 0). If you see it as a limit in Pr^R, every level is more or less the category of sheaves supported on B(0,1/n) and you are taking the intersection

Saal Hardali

Yeah that's what I was about to say

It looks just like vector spaces

Denis Nardin

Well, no. The category of sheaves supported at 0 is way bigger than just D(k)

Saal Hardali

7:39 PM

How is that?

Denis Nardin

I think at least.. this is one of the situations where you need to get the direction of the recollement right

Saal Hardali

Maybe I'm missing something but isn't there an exicision fiber sequence?

I can just take $j_!j^!$ and by the long exact sequence in cohomology i recover the sheaf

there's nothing except the $j^!\mathcal{F}$ where $j$ is the inclusion of the point

Denis Nardin

Hmm.. yeah I got confused by algebraic geometry where it's the other way round

Saal Hardali

haha

yeah

there you get something huge for sure

Denis Nardin

(i.e. k[e]/e^2 is a sheaf on A¹ supported at 0 but it's not a k-module)

Saal Hardali

7:41 PM

yeah I got you

That's a nice result lol

the stalk of the sheaves of sheaves is the category of stalks

not sterribly urprising though

sheaf of sheaves*

Is it a general thing that colimits in presentable categories are in some sense better?

Wait, if it's not presentable its horrible anyway, forget what i said.

Denis Nardin

Well, better is in the eye of the beholder

They are often the right thing though

Saal Hardali

Its definitely more presentable

Tim Campion

I look at it this way. Oftentimes the categories we consider are of the form $C = Ind(C_0)$, and we're really trying to understand $C_0$, while, the Ind part is there to make more things representable and so forth. When taking a colimit of presentable cateogries, we're often kind of more interested in a colimit of C_0's than of C's.

Denis Nardin

Compare the discussion in this chatroom a couple days ago about the difference between the two colimits of $\mathrm{Space}_*\xrightarrows{\Sigma}\mathrm{Space}_*\xrightarrow{\Sigma}\cdots‌$. The colimit in presentable categories is spectra. The colimit in categories is... well whatever it is. It's just not interesting

Tim Campion

The colimits in $Pres^L$ are such that $colim(Ind(C_0^i)) = Ind(colim(C_0^i)$ where the inside colimit is taken in finitely-cocomplete categories. So they allow us to compute such colimits using these right adjoint- limit type descriptions, giving us a payoff for passing to the Ind category.

Saal Hardali

7:50 PM

Good to know... Thanks!

I have an algebra question which might have a quick fix:

Let $A$ be a (classical) commutative ring and let $M$ and $N$ be modules over $A$. There's the following canonical morphism in the category of $A$-modules:

$$(\ast) \space \space \space Hom_A(M,A) \otimes_A N \to Hom_A(M,N)$$

One can ask: **When is this morphism an equivalence?**

It follows from Lazard theorem (on the characterization of flat modules) and from compact generation of the category that the answer is:

I mean is there a word (whenever we are in a symmetric monoidal stable category) and which if we replace it with flat the above theorem holds?

I mean the word is defined in every symmetric monoidal stable category not that the theorem holds in every such category.

Denis Nardin

Nevermind figured it out

Tim Campion

I think it's just "dualizable".

I asked a question that I think is related not long ago:

10

Q: What do you call $C$ if $[D,C] = D^\vee \otimes C$ for all $D$?

This is different from $C$ being dualizable ($[C,D] = C^\vee \otimes D$). But for example, if $C$ is a locally free sheaf of finite rank on a scheme/locally ringed space $X$, then $C$ has this property in the category of quasicoherent sheaves, or in the category of $\mathcal O_X$-modules -- just...

ct.category-theory monoidal-categories duality

I guess I was just asking in the non-infinity setting, but I imagine it generalizes.

Oh wait -- you want a condition on M, not on N

Denis Nardin

That is weird. Infinite free modules are not dualizable but it seems to me that the condition holds for them

Tim Campion

Actually, why?

Saal Hardali

Yeah that's what I was about to say. Lazard theorem is playing a highly non-trivial role here

Denis Nardin

7:57 PM

The argument in the answer to your question successfully argues that N dualizable => the condition holds, but I'll have to think about the other direction

Tim Campion

On the LHS, taking the colimit outside, it stays a colimit, but on the RHS, taking it outside it becomes a limit

right?

Saal Hardali

I'm confused about which problem we are talking about now

Tim Campion

Yeah, ignore the question I posted.

Denis Nardin

@TimCampion I think Saal is asking exactly the same question as you did in you MO question

The other thing (conditions on M) is well known to be equivalent to dualizability

Saal Hardali

could you explain in a bit more detail?

Denis Nardin

8:01 PM

I'm just saying that Hom(M,A)⊗N→Hom(M,N) is an isomorphism for all N iff M is dualizable

Saal Hardali

That's nice to know but that's not my question

in my question the variable is $M$

Denis Nardin

Exactly, you're asking the other thing: if that map is an isomorphism for all M, what can you say about N

Saal Hardali

Aha! I see what you're saying.

Yes this is my question.

Denis Nardin

And that is precisely the question Tim asked on MO. I'm a bit skeptical about the answer he got though

The answer says that if N is dualizable, then the map is an isomorphism for all M. This part of the proof seems sound

It's the other direction that leaves me perplexed

Saal Hardali

It must be wrong though no?

I mean an infinite sum is a counter example

Denis Nardin

8:05 PM

Ah wait, I'm not sure the infinite sum is indeed a counterexample

The problem is Hom(-,A)⊗N does not commute with colimits

(well, it doesn't turn colimits into limits)

So, I think that N is flat iff the map is an isomorphim for all finitely generated M, not for all M

Tim Campion

Maybe need finitely-presentable even?

Denis Nardin

While N is projective fg (i.e. dualizable) iff the map is an isomorphism for all M

Well, maybe finitely presentable is enough. If N is flat that map is an iso for all fg M though

Saal Hardali

If $N$ is free of infinite rank then you can take the sum outside and then no matter what you plug you get an infinite sum of it's dual. on the other side you get the same by presenting M as a colimit of compact sruff

Tim Campion

When you take the colimit presentation of M outside of Hom, it becomes a limit

Denis Nardin

Well, not you cannot present M as colimit of compact stuff because the lhs does not indeed turn filtered colimits to limits

Saal Hardali

8:08 PM

aha you're correct!

wait so as I stated it my question is wrong?

this is weird... I used it in practice several times....

Hopefully i did only for finite type $M$

Denis Nardin

As a bonus, you get another proof that flat +fg => projective

Saal Hardali

haha nice.

although it's probably circular since i have no idea what lazarad theorem uses

Are you sure about the only if in the new version?

Denis Nardin

Which of the two? The one for the flat case I haven't really checked

Saal Hardali

Aha wait nevermind got it

Just taking an infinite sum is enough

filtered limits don't commute with infinite sums that's the problem

cofiltered limits*

So it is true that $N$ is flat iff the original statement holds for all finitely presented $M$

What I thought originally is that one can turn this into a definition of flat in the derived setting... though better to ask if there are justifications for it...

Tim Campion

I guess my impression was that "going derived makes everything flat", and that in some sense "this is the whole point of going derived".

Denis Nardin

8:19 PM

The "classical" definition of flat in DAG is a filtered colimit of free modules (aping Lazard theorem). This turns out to be a very restrictive condition though. Unfortunately, the straightforward generalization of your condition (Hom(M,A)⊗N→Hom(M,N) is an equivalence for all compact M) is a wee bit too general (all N satisfy it!)

Saal Hardali

Oh right all $N$ satisfy this because dualizable=compact

What a disappointing conclusion...

Perhaps the opposite. It just always holds so i can just forget about this. less technical stuff

Tim Campion

Where does the condition "$\pi_\ast(M \otimes M \otimes M) = \pi_\ast(M \otimes M) \otimes_{\pi_\ast(M)} \pi_\ast(M \otimes M)$ fall? Somewhere between these two extremes?

Er -- whatever I just said, it was meant to be the condition that appears when identifying the E_2 page of the Adams spectral sequence

Denis Nardin

I'm not sure. When A is connective the first definition is equivalent to $\pi_0M$ is a flat $\pi_0A$-module and $\pi_iM=\pi_iA\otimes_{\pi_0A}\pi_0M$. This probably should imply your condition at the very least

Tim Campion

Here's what I meant.

Denis Nardin

Ah that one. I'm not sure. It feels like a completely orthogonal condition (it's there to ensure we have a Hopf algebroid instead of a more complicated simplicial scheme).

Tim Campion

8:41 PM

Right. So if $M$ is an $A$-module, we have a simplicial $A$-module $X_\bullet$ whose $n$th term is $A^{\otimes n-1} \otimes M$, and then a simplicial graded $\pi_0(A)$-module $\pi_\ast(X_\bullet)$. At least it's the case that "DAG-flat" and "ASS-flat" are both conditions on $\pi_\ast(X_\bullet)$. Somehow I want to say that "DAG-flat" is a "0-truncation" condition and "ASS-flat" is a "1-truncation" condition...

Transcript for