Homotopy Theory - 2017-01-15

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Joe Berner

01:17

I've given up on the previous question, so here's another: if I have a 1-categorical localization L : C \rightarrow D at some class of morphisms W, I can look at simplicial presheaves on both + projective model structure (Dugger's universal homotopy theory). Is sPre(D) in general Quillen equivalent to the Bousfield localization of sPre(C) at the class generated by W? C,D, and W are essentially small for my purposes.

Charles Rezk

@JoeBerner If you say $L$ is the $\infty$-categorical localization, rather than the $1$-categorical localization, the answer should be yes.

If I understand your question correctly

9 hours later…

Andrew Thompson

10:33

Are there non-affine schemes for which the structure sheaf generates the derived category of quasi-coherent sheaves?

2 hours later…

Arrow

12:12

Hello,
Is there some kind of conceptual or synthetic way to arrive at the definition of a derivation?

Joe Berner

12:52

@CharlesRezk The C and D I'm looking at are just 1-categories and L doesn't have any kind of adjoint. You can get a Quillen adjunction F: sPre(C)[W^{-1}] \rightarrow sPre(D) by universal properties, I guess the question is just if this is an equivalence. While L doesn't have a section, we can map D into sPre(C) via sending Y to Hom(L-,Y), and then postcompose to sPre(C)[W^{-1}]. Then by Dugger sPre(D) has a contractible space of Quillen functors to sPre(C)[W^{-1}].

We can pick one, call it G, and check that it's the 'inverse' to F via universal properties. (I think that makes sense, but the coffee hasn't kicked in yet.)

2 hours later…

skd

14:40

@Arrow these are just linear operators satisfying Liebniz. you can think of this in terms of tangent spaces. see this for example: mathoverflow.net/questions/88880/…

Joe Berner

15:06

@Arrow I'm a simple man and find K\"ahler differentials perfectly conceptual, but 7.3 in Lurie's Higher Algebra does translate the idea into categorical language so you can generalize this to \mathbb{E}_\infty rings

1 hour later…

Arrow

16:29

Thank you.
The [nlab entry](https://ncatlab.org/nlab/show/transversal+maps#properties) on tranversality says the tangent bundle functor preserves transverse pullbacks. How does this fit with the following example?

Let $f:X\to Y$ be a submersion and look at some fiber $f^{-1}(y)$. This is a transverse pullback. Its image by the tangent bundle functor is a commutative square (over $Y$) whose top left corner is the tangent bundle of the fiber $\mathrm Tf^{-1}(y)$. But $\mathrm T(\mathbf 1)=\mathbf 0$ and so if this square is also a pullback we conclude $\mathrm Tf^{-1}(y)\cong \mathrm{Ker}(\m…

3 hours later…

Denis Nardin

19:49

@Arrow The tangent bundle of the fiber is indeed the pullback of the kernel of $df$. I am not sure I understand your perplexity

Arrow

20:32

@DenisNardin doesn't that mean all fibers have the same tangent bundle? I thought that's strange since fibers of a submersion might not even have the same homotopy type

Mike Miller

see Ehresmann's theorem

Arrow

20:49

Ehresmann's theorem addresses proper submersions only though

I asked my question here

Mike Miller

21:01

Also, Ker(df) is that kernel restricted to each fiber. There's no reason that should imply the fibers are the same.

Denis Nardin

@Arrow You should precise in which category we are working. I think you should be working in the category of spaces equipped with a bundle, where the fiber products are computed by computing the fiber products of the underlying spaces, and then taking the fiber product of the pullbacks of the bundles

Arrow

21:25

@MikeMiller I'm not looking at the fiberwise kernel, I'm looking at the kernel bundle.

@DenisNardin I think the category of vector bundles over $X$ and then over $Y$ are the correct settings: we start by forming the kernel bundle in the category of bundles over $Y$ and then base change to obtain the vertical bundle, which is over $X$.

I tried to be detailed in my linked question :)

Denis Nardin

I do not participate to M.SE, but I cannot understand what you mean by "the tangent bundle functor"

The only such functor I can imagine goes from (smooth) manifolds to the category of spaces equipped with a vector bundle

Arrow

I'm thinking of the codomain simply as the category of vector bundles

The internal vector spaces in the arrow category of manifolds?

Denis Nardin

What is the category of vector bundles?

Arrow

Does that not make sense?

Denis Nardin

I think that you mean the same thing as I mean when I say "the category of manifolds equipped with a vector bundle", where objects are pairs (M,E) where M is a manifold and E is a vector bundle over M and morphism are pairs of maps (f:M→M',E→f^*E')

Arrow

21:30

Oh, yes, sorry.

Denis Nardin

Ok, then what are fiber products in this category?

That is suppose you have a diagram (M',E')→(M,E)←(M'',E''), what is the pullback?

Arrow

I'm not looking at pullbacks within this category, but in the category of manifolds. Each submersion gives a base change functor

Denis Nardin

I think you are confusing two different functors, or perhaps I do not understand what you mean

I thought that you were asking about how T:Mfld→(Mfld with a v. bundle) preserves pullbacks

Arrow

Ahh, I understand where I've mislead you. Sorry. I am just trying to understand why $T$ from manifolds to manifolds preserves pullbacks. So, compose with forgetting about the bundle structure in the codomain.

A manifold goes to the total space of its tangent bundle

If I understand correctly, that's what the nlab says here

Denis Nardin

Actually I'd intepret it as saying that the functor T I described above preserves pullbacks

Arrow

21:35

Ok, I will try to think about the functor you described earlier later, but for now I don't understand how the latter could be true

Should I restate what's bugging me?

Denis Nardin

Sure

Arrow

Start with a pullback square $$\require{AMScd} \begin{CD} f^{-1}(y) @>>> \mathbf 1\\ @VVV @VV{y}V\\ X @>{f}>> Y \end{CD}$$. Apply $\mathrm T$ to obtain a commutative square (over $Y$) $$\require{AMScd} \begin{CD} \mathrm T f^{-1}(y) @>>> \mathbf 0\\ @VVV @VVV\\ \mathrm T X @>{\mathrm df}>> \mathrm T Y \end{CD}$$. If this square is a pullback, the left vertical arrow is the kernel-bundle of the differential.

Denis Nardin

Well, wait. The square is a pullback in what category?

Arrow

The latter? Maybe this is where I got confused... I was thinking of its as a pullback square in the category of manifolds...

Denis Nardin

The kernel bundle of the differential is the pullback of the square TX→f^*TY←0 in the category of vector bundles over X, but I thought we were just working in the category of manifolds here

Then the pullback is not the kernel bundle of the differential

In fact, 0→TY is not the zero section but the inclusion of the zero vector above y

Arrow

21:40

I don't understand your last sentence - it's just the initial arrow, no?

Denis Nardin

No

So, instead of calling it 0, let me call it T1

It is also the 1-point space

That is not the initial object in the category of smooth manifolds, the initial object is the empty manifold

Arrow

I kept confusing two categories because I was also thinking of the category of vector bundles, which has a zero object

Bleh, I mixed everything up.

If we replace my incorrect zero with the terminal object as you said (I undertand what you meant by the zero vector above $y$ now), then the square $$\require{AMScd} \begin{CD} \mathrm T f^{-1}(y) @>>> \mathbf 1\\ @VVV @VVV\\ \mathrm T X @>{\mathrm df}>> \mathrm T Y \end{CD}$$ should be a pullback, right?
(In the category of manifolds)

Denis Nardin

Yes, and in fact it is

Arrow

Ahh, okay. I think it all makes sense now, since the right vertical arrow does not become just zero, but remembers the point $y$.

Denis Nardin

You can easily deduce that this square is a pullback from the implicit function theorem

Arrow

21:50

I definitely can't do it myself but the books I'm looking at deduce the fiberwise result

Denis Nardin

You should try to do it yourself, it's a good exercise in differential geometry and it is not that hard

I found that a bit of hands on experience is invaluable if one is trying to tackle abstract subjects

Arrow

That is, first prove fibers of regular values are embedded premanifolds, then use the local form of submersions to deduce the kernel of the differential at a point is the tangent space to the fiber at the same point.

Denis Nardin

Yeah, you can even get it in one step if you word your argument carefully

Arrow

Ok. Is it okay if I ask for a solution in a couple of days?

Denis Nardin

Sure. Why wouldn't it be?

Arrow

21:52

I'm wasting plenty of your time :)

Denis Nardin

Well, I do not need to be the one to answer...

Arrow

Last thing that's been bugging me is the kernel bundle. In particular, its fibers. By definition the kernel bundle is over $Y$, so $(\ker \mathrm df)^{-1}(y)\overset{?}{=}\varinjlim_{p\in f^{-1}(y)}\ker\mathrm d_pf$ in the category of vector spaces?

Denis Nardin

The kernel bundle is not over Y. Why do you think it should be over Y?

The kernel bundle is over X

Arrow

I thought it's over $Y$ because kernels do not make sense in a category of vector bundles unless the base is fixed

Denis Nardin

Indeed, it is the kernel of TX→f^*TY

Arrow

21:56

The kernel bundle is over $Y$, but the vertical bundle is the base change of the kernel bundle along $f$. Right?

Denis Nardin

Which definition of kernel bundle are you using?

Arrow

the bundle with the universal property of the kernel in the category of vector bundles over $Y$

Denis Nardin

But the kernel of what map?

TX→f^*TY is a map of vector bundles over X

Arrow

I'm thinking of the kernel bundle as the kernel of the differential of $f$, seen as a bundle map $df:f\circ \pi\to \pi^\prime $ where $\pi:TX\to X$ and $\pi^\prime :TY\to Y$.

Denis Nardin

What is $f\circ π$?

Arrow

21:59

the composite $TX\to X\to Y$

Denis Nardin

That is not a vector bundle over Y

I mean, its fibers are not vector spaces

Arrow

Yes, that bugged me, that's why I started fabricating stuff with colimits..

Denis Nardin

It's much better thinking of df as a map from TX to f^*TY

So, a map of vector bundles over X

Arrow

because then it really is a vector bundle map of bundles over $X$?

Denis Nardin

Yes

Then its kernel is really what's called the "vertical bundle", that is the vector bundle over X which is the subbundle of TX consisting of vectors tangent to the fibers of f

Arrow

22:02

I understand! Do you ever think of $df$ as going into $TY$, or always into the base change?

Denis Nardin

Well, it depends of what I am using df for. In general I think of df as a map in the category I outlined above (topological spaces equipped with a vector bundle) which then can take a particular aspect depending on what characteristic I'm interested in

Arrow

But in the category you mention, the differential is really a commutative square whose top edge is $df:TX\to TY$, right?

Denis Nardin

It depends on the precise description of the category. An equivalent description is saying that a map (X,E)→(X',E') is a pair of maps (f,\tilde f) where f:X→Y is a map of spaces and \tilde f: E→f^*E' is a map of vector bundles over X

Arrow

Okay. One last question: are the surjective submersions by any chance precisely the regular epis of the category of (pre)manifolds? (maybe the universally regular epis?)

Denis Nardin

Uhm I do not think so, but I am far from an expert.

That is quite orthogonal to what submersions are good for (the existence of well-behaved pullbacks)

Arrow

22:09

I just thought the embeddings would be the regular monos and surjective submersions the regular epis because it would give a neat picture

Denis Nardin

It might be the case, but even if it is true I do not think it's going to be anything but an interesting curio

Arrow

Okay, thank you very very much for everything! I'm off to bed.

Denis Nardin

Good night!

Mingcong Zeng

22:52

What do we know about Picard group of homotopy category of G-spectra for G a cyclic group of order p^n? Do we have a explicit description of it somewhere?

Denis Nardin

@Mingcong Yes! There is a complete description of Pic(G) for every finite group here: math.uiuc.edu/K-theory/0407

To find the complete statement in full generality you have to hunt down a few references though

I think that in the case of a p-group it might be completely generated by the representation spheres plus the picard group of the Burnside ring

Mingcong Zeng

23:11

@DenisNardin Awesome! I am reading that paper now! Thank you!

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