Large Smooth Categories

June 8, 2007

Posted by Urs Schreiber

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Behind the scenes, I am having a long email discussion with Bruce Bartlett about some puzzling subtleties in the definition of smooth categories.

As Bruce rightly emphasized, we need to be careful with comparing the following two definitions

1) a category internal to the category of Chen-smooth spaces

2) a stack over manifolds – which is the same as a category fibered over manifolds.

It gets particularly subtle when the categories in question are large. I am unsure about some subtle details of that. Here are some questions.

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Reminder: Chen-smooth spaces (Diffeology).

For my purposes here, a “smooth space” is a special sort of sheaf (of sets) on manifolds, namely a sheaf $X$ such that for each manifold $U$ the set $X (U)$ is a subset of the set of maps-of-sets $U \to X,$ where $X$ is any fixed set.

Such a sheaf is known as a Chen-smooth structure on the set $X$ . (See Quantization and Cohomology (Week 20)). In this context the elements of $X (U)$ are called plots of $X$ .

A puzzling example: Internal smooth structure on $Vect$ ?

In our discussion, I imagined realizing the category of vector spaces as a category internal to smooth spaces by defining the following sheaf:

For each manifold $U$ , the set $Mor (Vect) (U)$ is that of maps $U \to Mor (Vect)$ which arise as component maps of smooth morphisms $\begin{matrix} V_{1} & \overset{ϕ}{\to} & V_{2} \\ ↘ & ↙ \\ U \end{matrix}$ of smooth vector bundles $V_{1} \to U$ and $V_{2} \to U$ over $U$ .

I did notice that this sounds like there might be a subtlety hidden. Bruce very much emphasized that we need to be careful concerning sets and classes here. I realize I don’t have this sufficiently under control, conceptually.

Does the above definition of a sheaf of plots make sense? If not, is there any chance of fixing it, say by mumbling something about Grothendieck universes? Whatever the answer is, I would very much appreciate a detailed explanation.

A non-puzzling example: $Vect$ as a smooth stack.

Alternatively, we may equip $Vect$ with a notion of smooth structure by realizing it as the value over the point of the standard stack $Vect$ which sends manifolds to the category of smooth vector bundles over them.

This should be the same, dually, as a fibred category: thinking of $Vect$ as the category of smooth vector bundles over all possible manifolds (morphisms are morphisms of bases spaces together with morphisms of of one bundle with the pullback of the other), together with the obvious forgetful functor $Vect \to Manifolds,$ which is a fibred category.

Questions.

Before I completely abandon the idea of being able to realize $Vect$ internal to Chen-smooth spaces, I would like to understand precisely what is going on.

How do I decide if $Mor (C)$ is a set or a class? If it is a class, are maps from sets into it a set or a class? Is there any chance to get a sheaf (of sets) of maps into $Mor (Vect)$ ?

The issue in full generality.

What I really would like to understand one fine day (of course Bruce already made a couple of remarks on that) is how the following two concepts are related, generally, for $S$ any site:

A) categories internal to sheaves over $S$

B1) categories fibred over $S$

B2) stacks on $S$

Can anyone educate me here? Thanks!

Posted at June 8, 2007 12:51 PM UTC

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Re: Large Smooth Categories

I am confused. Is Vect a stack? For example, Vect(point) doesn’t look like a groupoid. Nor do I see how the descent would work. It’s true that Vect^n, the category of vector bundles of a fixed rank n is a nice Artin stack, but Vect?

Posted by: Eugene Lerman on June 8, 2007 4:24 PM | Permalink | Reply to this

Re: Large Smooth Categories

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I am confused.

It’s probably me who is confused and spouting nonsense!

$Vect (point)$ doesn’t look like a groupoid.

I was thinking of stacks with values in $Cat$ instead of just in $Grpd$ , since the idea was that we think of a smooth category (not necessarily a groupoid) as a category whose objects are families parameterized smoothly over manifolds.

Nor do I see how descent would work.

Oh, I see, descent depends on the fact that morphism are invertible, right.

Okay, so maybe it’s not a stack.

It’s true that ${Vect}^{n}$ , the category of vector bundles of a fixed rank $n$ is a nice Artin stack

If we restric to isomorphisms of bundles?

Hm, all right. But what about fibred vategories. Do arbitrary vector bundles over manifolds (with arbitrary morphisms between them, the way I indicated), a fibred category $Vect \to manifolds$ ?

Posted by: urs on June 8, 2007 5:28 PM | Permalink | Reply to this

Re: Large Smooth Categories

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Another possibility is that actually I shouldn’t be asking about stacks, but just about prestacks. Yeah, right, that’s actually closer to the original question.

So it could be that what I really need to know is what the answer to my above questions is with “stack” everywhere replaced by “prestack” (i.e. just any pseudofunctor $manifolds \to Cat$ ).

But anyway, all replies, to whichever flavor of this problem, are very welcome.

Posted by: urs on June 8, 2007 5:31 PM | Permalink | Reply to this

Re: Large Smooth Categories

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A paper that’s perhaps of relevance here is Differentiable Stacks and Gerbes by Behrend and Xu. This paper defines a differentiable stack as an ordinary stack $F$ on $Man$ together with a representable surjective submersion

(1)

π : X \to F

from a Hausdorff smooth manifold $X$ into $F$ , called an atlas .

Perhaps this idea fits in well with the notion of anafunctors, where coverings and surjective submersions seem to play a similar role.

Posted by: Bruce Bartlett on June 9, 2007 7:24 PM | Permalink | Reply to this

Re: Large Smooth Categories

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Perhaps this idea fits in well with the notion of anafunctors, where coverings and surjective submersions seem to play a similar role.

True, but right now this doesn’t seem to help answering our question here, or does it?

As I mentioned above, possibly stacks here were a red herring. Instead we ought to be looking at categories fibered over manifolds.

But most urgently, I would still like to better understand whether it makes sense to equip $Vect$ with a Chen-smooth structure as indicated in the entry above, or not.

Somehow there must be a relation between

- categories internal to sheaves over manifolds

and

- categories fibered over manifolds.

Which is it?

Posted by: urs on June 10, 2007 4:07 PM | Permalink | Reply to this

Re: Large Smooth Categories

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It is true that the bicategory of smooth stacks in groupoids is equivalent to the bicategory of smooth groupoids and anafunctors (and some notion of transformation I can’t remember at the moment) This was showed by Pronk in the mid-90’s. I expect something similar to be true for categories as opposed to groupoids so assume this result. We have an embedding

$Man ↪ Shv (Man)$

and so can consider categories internal to $Man$ as representable sheaves. This categories fibred over $Man$ can be considered as categories internal to (representable) sheaves. I reckon it is the case that can reverse this process, and the “representable internal categories” are precisely the fibred categories over $Man$ .

On another point, let $X$ be our smooth space (a set with specified plots). Then the Chen smooth structure on $X$ generates a sieve on $X$ as an object of $Set$ . Given some notion of Grothendieck topology on $Set$ it would be nice to use this sieve to get a topology on $X$ (or $Set / X$ , but this I think less likely.) Is there an automatic notion of topology on $Set$ ? Presumably covering families are just collections ${X_{α} \to X}$ such that $\cup_{α} X_{α} \to X$ is onto.

Posted by: David Roberts on June 12, 2007 6:35 AM | Permalink | Reply to this

Re: Large Smooth Categories

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It is true that the bicategory of smooth stacks in groupoids is equivalent to the bicategory of smooth groupoids and anafunctors

Do we have to restrict to saturated anafunctors? That’s at least what Toby Bartels seemed to indicate, since these saturated anafunctors between groupoids are supposed the same as Hilsum-Skandalis morphisms of groupoids, which one may think of as certain bitorsors of groupoids.

By the way, it is probably clear to you, but the fact you mention became fully intuitively clear to me only after I learned that a stack “presented” by a Lie groupoid $G$ is nothing but the stack of $G$ -principal groupoid bundles, i.e. that of $G$ -torsors. That nicely explains the situation: a Morita morphism of gadgets (here: Hilsum-Skandalis morphism of groupoids) corresponds to a morphism of modules for that gadget (here: groupoid torsors).

This categories fibred over Man can be considered as categories internal to (representable) sheaves.

Ah, that’s what I am looking for! How can I see this?

let $X$ be our smooth space (a set with specified plots)

What if I want $X = Mor (Vect)$ ? Any chance to make sense of that?

Posted by: urs on June 12, 2007 9:56 AM | Permalink | Reply to this

Re: Large Smooth Categories

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Thus categories fibered over $Man$ can be considered as categories internal to (representable) sheaves.

This is the part I don’t understand. The point is that categories internal to (representable) sheaves must necessarily have a set of objects and a set of morphisms, by definition.

But that eliminates many of the interesting examples of categories fibered over manifolds! For brevity we’ll refer to a category fibered over Man as a prestack, i.e. a weak 2-functor ${Man}^{op} \to Cat$ .

For instance, the vector bundles prestack ${Vect}_{s}$ . It associates to $U \in Man$ the category ${Vect}_{s} (U)$ of vector bundles over $U$ . But it seems that ${Vect}_{s} (U)$ doesn’t have a set of objects or a set of morphisms.

Or the cobordism stack, say $2 {Cob}_{s} (X)$ . It associates to $U \in Man$ the category $2 {Cob}_{s} (X) (U)$ whose objects are “smooth families of closed 1d-manifolds mapped into $X$ ” and whose morphisms are “smooth families of cobordisms mapped into $X$ ”. These things don’t form sets. Unless we start choosing small models or start using Grothendieck universes… which are things I don’t understand.

To conclude, intuitively it seems to me that to make this comparison exact what one would need is the concept of a “classy sheaf”, i.e. a functor

(1)

{Man}^{op} \to Classes

where “ $Classes$ ” is the “category of classes”. To my knowledge though, such a category doesn’t make sense. (Or does it? See P.S.)

In short, I feel that it is better to think of

(2)

a collection of categories {C_{U}}, one for each manifold U

than to think of

(3)

a collection of sets of objects {{Ob}_{U}}, and a collection of sets of morphisms {{Mor}_{U}}, one for each manifold U .

P.S. Could someone explain to me why the category of classes doesn’t make sense? (If indeed it doesn’t.) There are two possible objections : (a) the “collection of all classes doesn’t make sense” and (b) perhaps “functions between classes don’t make sense”.

But if objection (a) is true, then it would also cancel out Cat, the 2-category of categories, since the collection of objects of Cat are similar to the “class of all classes”. And if (b) is true, then it would also cancel out the possibility of a functor between categories! Toby… can you help?

Posted by: Bruce Bartlett on June 12, 2007 10:41 AM | Permalink | Reply to this

Re: Large Smooth Categories

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Saturated anafunctors? Probably. The Pronk theorem actually uses generalised morphisms, which are spans where the left leg is a fully faithful, essentially surjective smooth functor, and shows we can localise ff ess. surj. functors. I never quite got the reason for talking about anafunctors and then talking about saturated ones. I think in my book anafunctors are automatically saturated.

When I say

Thus categories fibered over Man can be considered as categories internal to (representable) sheaves

I mean with saturated anafunctors as morphisms.

Posted by: David Roberts on June 14, 2007 5:38 AM | Permalink | Reply to this

Re: Large Smooth Categories

Actually, can we back up the truck just a tad? I’ve only been getting into anafunctors at all very recently. Though I’m finding them incredibly useful in my own ways, I’m still very much an initiate.

So, that said: what’s a “saturated” anafunctor?

Posted by: John Armstrong on June 14, 2007 6:17 AM | Permalink | Reply to this

Re: Large Smooth Categories

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I think Toby’s thesis is the best place to start - he calls it a 2-map though. I found Makkai’s account a little confusing, not to mention his typesetting (sorry, but TeX has spoilt me).

But from what I understand, is that a saturated anafunctor

$F : X \overset{\cdot}{\to} Y$

It’s usually expressed in more `internal’ language though. is a fully faithful surjective functor $f_{1} : \tilde{X} \to X$ and a functor $f_{2} : \tilde{X} \to Y$ . Please someone correct me if I’m wrong

Posted by: David Roberts on June 14, 2007 7:48 AM | Permalink | Reply to this

Re: Large Smooth Categories

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I think what David gives is the definition of an ordinary anafunctor. A saturated anafunctor satisfies one extra condition.

So, as David Roberts says, an anafunctor

$F : X \overset{\cdot}{\to} Y$

is a span of ordinary functors $\begin{matrix} ∣ F ∣ & \overset{f}{\to} & Y \\ ↓^{p} \\ X \end{matrix}$

such that $p$ is, morally, a surjective equivalence of categories, but we just say that it is a fully faithful functor, surjective on objects, since one of the main points of this business is that we do not want to require that a weak inverse of $p$ actually exists internal to the context we are working on.

(Makkai put himself in the context of the category of sets without the axiom of choice. That implies that weak inverses of $p$ usually do not exist, since constructing them would amount to finding sections of $p$ on objects.

Toby Bartels formulated everything more generally in internal language. His motivation is the context of smooth categories. Here the weak inverse does not exist whenever $p$ is, on objects, a nontrivializable smooth bundle.)

Next, an anafunctor as above is called saturated precisely if, for all $x \in Obj (X)$

$f (p^{- 1} ({Id}_{x}))$

fully exhausts one isomorphism class of objects.

(Here $p^{- 1}$ denotes the preimage, not the inverse!)

See p. 11 of Makkai’s text.

(There must be a better way to say this, though, I guess…)

A while ago I here had a discussion with Toby about saturated anafunctors and Hilsum-Skandalis morphisms starting here.

Posted by: urs on June 14, 2007 10:37 AM | Permalink | Reply to this

Re: Large Smooth Categories

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David Roberts wrote:

When I say

Thus categories fibered over Man can be considered as categories internal to (representable) sheaves

I mean with saturated anafunctors as morphisms.

Please bear with me, I must be missing something basic: let’s not worry about morphisms for a second, I haven’t even understood this statement on objects, yet.

Do you really mean to restric to representable sheaves over, say, $S$ ? Then that would be like saying that categories fibered over $S$ are equivalent to categories internal to $S$ itself. No? But that can’t be right, can it?

I am probably confused. Could you maybe help me by spelling out how you imagine constructing from a fibered category an internal one, and vice versa?

Posted by: urs on June 14, 2007 10:45 AM | Permalink | Reply to this

Re: Large Smooth Categories

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To get from an internal cat $C$ (in $D$ ) to a fibred cat, one just takes the fibred cat

${Hom}_{D} (-, C)$

of internal functors etc. The stackification of this is the stack of $C$ -torsors (I don’t know how well this goes for cats as opposed to groupoids).

The other way, I don’t know if it works for fibred cats, but for stacks in groupoids on $D$ , one takes a presentation/atlas/chart (a map from a representable stack to the one in question, the map satisfying some representability conditions) and then the groupoid $A$ coming from this is internal to $D$ . But $A$ is only unique up to equivalence, and only then when we use (saturated) anafunctors.

The point is that the firbed cat itself isn’t internal, but one can get something in an equivalent category which is.

Posted by: David Roberts on June 15, 2007 2:28 AM | Permalink | Reply to this

Re: Large Smooth Categories

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To get from an internal cat $C$ (in $D$ ) to a fibred cat, one just takes the fibred cat

${Hom}_{D} (-, C)$

of internal functors, etc.

All right. I am looking for a pseudofunctor on $D$ (not a 2-functor on categories internal to $D$ ). So let’s see, I guess the point is we can do the following:

We consider the category ${DiscCat}_{D} \subset {Cat}_{D}$ of all those categories internal to $D$ which are discrete in that they only have identity morphisms.

Then, using $D ≃ {DiscCat}_{D}$ we restrict ${Hom}_{{Cat}_{D}} (-, C) : {Cat}_{D} \to Cat$ along the inclusion $D \subset {Cat}_{D}$ as above and obtain a pseudofunctor ${Hom}_{{Cat}_{D}} (-, C) : D \to Cat .$

And if we look at what this looks like in detail, we find indeed that it is the kind of thing we were talking about:

to any manifold, say, $U \in D$ this functor associates the category whose

objects are $U$ -parameterized families of objects in $C$ ;

morphisms are $U$ -parameterized families of morphisms in $C$ .

Okay, very nice!

Now, the main issue was to what extent this is onto. Does an inverse to this operation exist, or are categories fibered over $D$ intrinsically more general than categories internal to $D$ ?

The other way, I don’t know if it works for fibred cats, but for stacks in groupoids on $D$ , one takes a presentation/atlas/chart (a map from a representable stack to the one in question, the map satisfying some representability conditions) and then the groupoid $A$ coming from this is internal to $D$ . But $A$ is only unique up to equivalence, and only then when we use (saturated) anafunctors.

Ah, I see! Thanks!

So maybe it’s that condition on representability of the atlas

$X \to our Stack$

which makes this work?

Let me see: my understanding is that a stack which is presented by a groupoid $G$ is (up to equivalence, I presume) the stack of principal $G$ groupoid bundles. (Right?)

I guess that’s the groupoid we are talking about?

Hm, now say I start with a Lie groupoid $G$ but am interested in regarding the category $G Tor$ (torsors over a point, i.e. objects are $G$ -spaces which are isomorphic to $G$ as $G$ -spaces) as internal to sheaves over smooth spaces.

If I understand correctly, by your prescription the internal category we get from the stack of $G$ -bundles back the groupoid $G$ itself. Which is internal to manifolds even.

Now, what I was looking for is a way to realize $G Tor$ as internal to sheaves over manifolds.

Do we get this from what you said? Probably the point is that while $G$ and $G Tor$ are not equivalent using internal functors, they do becomes equivalent using internal anafunctors. I guess that’s your point?

Posted by: urs on June 15, 2007 10:24 AM | Permalink | Reply to this

Re: Large Smooth Categories

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David Roberts makes a good point, and it has removed some of my misconceptions about stacks. I think this is basically Proposition 70 of Metzler (or better yet, proposition 77) : smooth groupoids basically correspond to locally representable stacks. Stack morphisms basically correspond to (saturated?) anafunctors between the corresponding smooth groupoids.

I have three questions :

1. What is a nice example of a non-locally representable stack?

2. In this context, how should we understand the “d-dimensional cobordism stack in $X$ ” - which is perhaps better viewed as the “d-dimensional cobordism category in $X$ fibered over manifolds”? Recall this is the one whose fiber at a manifold $U \in Man$ is the category whose objects are smooth families of $(d - 1)$ -dimensional closed manifolds in $X$ , parametrized by $U$ , and similarly whose morphisms are smooth families of $d$ -dimensional cobordisms in $X$ , parametrized by $U$ .

Is this locally representable, in a suitable sense?

3. So it seems to me that in the picture where locally representable stacks correspond to smooth groupoids, the stacky picture is the global one and the smooth groupoid picture is the “local” one. At least, that’s how it works for $G$ -bundles. Does that sound right or is it a bit misleading?

What Urs and I were discussing was seemingly something slightly different. We were debating in what situations it was possible to regard smooth groupoids (i.e. internal groupoids) also as global beasts; i.e. we were trying to contrast them on the same ontological level (if that makes any sense), not on the level where one is a “local presentation” of the other.

Posted by: Bruce Bartlett on June 15, 2007 12:45 PM | Permalink | Reply to this

Re: Large Smooth Categories

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I am still not sure if this answers the question which we actually fought with:

it wasn’t so much categories just internal to manifolds which we wanted to relate to categories fibered over manifolds, but categories internal to sheaves over manifolds.

The subtlety somehow rests in the fact that as mere categories internal to sets we have

$Σ G ≃ G Tor$

(on the left the category with a single object and $G$ worth of morphisms, on the right the category whose objects are $G$ -spaces isomorphic to $G$ and morphisms homomorphisms of these).

This equivalence works by simply choosing for each $G$ -space which is isomorphic to $G$ one particular isomorphism.

The issue is that this identification is a priori not smooth in any sense.

It seems to me that at the moment we are talking about how to think of $Σ G$ as a stack.

But the real issue we were trying to deal with was how to think of $G Tor$ as a category internal to sheaves over manifolds – and if that can somehow be canonically obtained from the obvious stacky realization of $G Tor$ .

Posted by: urs on June 15, 2007 1:17 PM | Permalink | Reply to this

Re: Large Smooth Categories

ΣG → GTor is a cofibrant replacement.

Running out of time on this computer…

Posted by: David R on June 16, 2007 7:30 AM | Permalink | Reply to this

Re: Large Smooth Categories

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Urs wrote:

The subtlety somehow rests in the fact that as mere categories internal to sets we have

$Σ G ≃ G Tor$

(on the left the category with a single object and $G$ worth of morphisms, on the right the category whose objects are $G$ -spaces isomorphic to $G$ and morphisms homomorphisms of these).

This equivalence works by simply choosing for each $G$ -space which is isomorphic to $G$ one particular isomorphism.

The issue is that this identification is a priori not smooth in any sense.

I think it actually is, but maybe you can check.

Suppose $G$ is any Lie group.

On the one hand, $Σ G$ is a smooth category in an obvious way: its smooth space of objects is the one-point space, while its smooth space of morphisms is $G$ .

On the other hand, $G Tor$ becomes a smooth category in more or less the same way that $Vect$ does. Namely:

To define the smooth space of objects of $G Tor$ , we just give the set of all $G$ -torsors its discrete smooth structure.

To define the smooth space of morphisms of $G Tor$ , first consider any pair of $G$ -torsors $X$ and $Y$ . The set $hom (X, Y)$ has a bunch of obvious bijections

$hom (X, Y) ≅ G$

Choose any one of these and use it to transfer the smooth structure on $G$ to $hom (X, Y)$ ; the result doesn’t depend on your choice. Next, note that the set of all morphisms in $G Tor$ is the disjoint union of these sets $hom (X, Y)$ . So, make the set of all morphisms into a smooth space by taking the coproduct of all the smooth spaces $hom (X, Y)$ .

Now for the punchline. Suppose we pick an isomorphism between each $G$ -torsor and $G$ itself. Then we get an equivalence

$A : G Tor \to Σ G$

$B : Σ G \to G Tor$

I claim that these functors are both smooth! And, I claim there are smooth natural isomorphisms

$α : B \circ A \Rightarrow 1_{G Tor}$

and

$β : A \circ B \Rightarrow 1_{Σ G}$

If I’m right, this means that $G Tor$ and $Σ G$ are equivalent in the 2-category

$[smooth categories, smooth functors, smooth natural transformations]$

In some ways it’s better to use a different 2-category when stating this result, namely

$2 C^{∞} = [smooth categories, smooth anafunctors, smooth ananatural transformations]$

I don’t think anafunctors and ananatural transformations are needed for the result to hold! Still, it’s worthwhile checking that $G Tor$ and $Σ G$ are also equivalent in $2 C^{∞}$ .

One reason is this. We already know that for any smooth space $X$ ,

$hom (P X, Σ G) ≃$ $[smooth principal G - bundles with connection over X, isomorphisms thereof]$

where $hom (P X, Σ G)$ is defined in $2 C^{∞}$ . In other words, $hom (P X, Σ G)$ is the category with smooth anafunctors

$F : P X \to Σ G$

as objects, and smooth ananatural transformations between these as morphisms.

If we indeed have

$Σ G ≃ G Tor$

in $2 C^{∞}$ , we can instantly conclude that we also have

$hom (P X, G Tor) ≃$ $[smooth principal G - bundles with connection over X, isomorphisms thereof]$

where $hom (P X, G Tor)$ is again defined in $2 C^{∞}$ .

If everything I’m saying is true, $G Tor$ and $Σ G$ are just different ways of talking about the same thing — not just as categories, but also as smooth categories!

Again, I’d be very happy if you’d would check this stuff… it seems right to me but I could be overlooking something. I really need these results, and my previously stated ones about vector spaces, to feel content with my philosophy of smooth categories.

Posted by: John Baez on June 16, 2007 9:47 PM | Permalink | Reply to this

Re: Large Smooth Categories

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Again, I’d be very happy if you’d would check this stuff

Don’t have time right now for thinking about it in detail, but I certainly do expect that this is right.

However, it is not quite what I was looking for!

What I am looking for (not necessarily, but sufficiently ;-) is a Chen-smooth structure on $G Tor$ such that ordinary Chen smooth functors

$P X \to G Tor$

are equivalent to smooth $G$ -bundles with connection.

If we use anafunctors, we can just as well stick with $Σ G$ as a model for $G Tor$ or do what you just sketched.

But the main motivation for all this here is to understand the relation of anafunctors (or things closely related to them) which sort of define smoothness in terms of “local models” to attemtps to define “globally smooth” functors.

For that to work, we really need a non-discrete smooth structure already on the objects of $G Tor$ .

It’s obvious how this works in the world of stacks. What I am trying to find out is if there is also a way to make it work in the world of categories internal to Chen-smooth spaces.

And, by the way, I now think the answer is: yes.

Posted by: urs on June 17, 2007 6:45 PM | Permalink | Reply to this

Re: Large Smooth Categories

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Urs wrote:

What I am looking for (not necessarily, but sufficiently ;-) is a Chen-smooth structure on $G Tor$ such that ordinary Chen smooth functors

$P X \to G Tor$

are equivalent to smooth $G$ -bundles with connection.

Oh! Wow!

I’m afraid I haven’t had time to follow your battle with Bruce. I had no idea what you were talking about… it sounded too complicated for a simple man like me. So, I only jumped in when you raised the issue of ‘set-theoretic difficulties’, which is something I understand. Then you asked me a question. But, not understanding the context, I guess I didn’t understand the question!

So, sorry. Anyway:

It seems rather bizarre to seek a smooth structure on $G Tor$ such that ordinary Chen smooth functors

$P X \to G Tor$

are equivalent to smooth $G$ -bundles with connection. It would be cool if one existed, but I can barely believe my eyes when you claim one does! So, I’ll ask:

Are you actually claiming that with some smooth structure on $G Tor$ , we have

$hom (P X, G Tor) ≃ [smooth principal G - bundles with connection over X, isomorphisms thereof]$

where the $hom$ is defined using the 2-category

$C^{∞} Cat = [smooth categories, smooth functors, smooth natural transformations] ?$

It’s hard for me to tell if this is really what you’re talking about in the comment you pointed me to. In particular, I’m having trouble finding your smooth structure on objects of $G Tor$ .

Personally, I’ve given up trying to use smooth functors except as a cheap shortcut here and there. I’ve decided that the really good thing is anafunctors. But, the half-proved theorems I stated here and here are necessary conditions for me to be happy with the anafunctor approach.

Posted by: John Baez on June 17, 2007 8:11 PM | Permalink | Reply to this

Re: Large Smooth Categories

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Hi John,

only have time for a very quick reply:

yes, I expect that smooth functors from paths to $G Tor$ , with everything internal to Chen smooth spaces, as indicated, are equivalent to $G$ -principal bundles with connection. I’d think this is even easy to prove, using the technology in my paper with Konrad. In fact, I skecthed part opf the proof already, albeit with a slightly too naive version of the smooth structure on $G Tor$ in mind.

But – I haven’t sat down yet and tried to prove it. I stopped worrying at the point where I got convinced that the “obvious” Chen-smooth structure on $G Tor$ actually does make sense. Bruce had to chase me quite a bit before I had formulated my original proposal in a way that actually makes good sense.

I think everything is easiest to understand if we forget connections for a moment and just consider bundles.

So, we are looking for smooth functors from $Disc (X)$ to $G Tor$ . As we mentioned somewhere, in the world of stacks it works precisely the way as expected:

a smooth $G$ -bundle is a morphism from the stack given by the space $X$ to the stack of $G$ -bundles – which is nothing but the stacky version of $G Tor$ .

My whole motivation for this discussion here was the question if we can replace stacks over manifolds here with categories internal to sheaves over smooth manifolds, and in particular with categories internal to Chen-smooth spaces.

For that to work, one has to manage to arrange such that something like $G$ -bundles over manifolds behaves like a sheaf instead of like a stack.

But that is precisely what one obtains after passing from bundles-as-total-spaces to bundles-as-fiber-assigning-functors.

So, that is my proposal now: use the way to conceive bundles as functors which I describe with Konrad to construct the idea of the stack $G Tor$ as a category internal to Chen-smooth spaces.

In particular, I’m having trouble finding your smooth structure on objects of GTor.

But this just follows from restricting to identity morphisms. Here, the sheaf of objects assigns to a manifold $U$ the collection of $G$ -bundles over $U$ . But this collection is conceived now as a colleciton of functors with a certain property. These do form a sheaf.

I have to go to bed now and from tomorrow on I’ll be on a conference. But after that I can try to write this up cleanly and comprehensively.

So, to clarify: I think I now have the answer to one of my main puzzles. But one thing we are still discussing here was if there is a general abstract nonsense (as opposed to my concrete construction above) which would allow one to pass from stacks over $S$ to categories internal to sheaves over $S$ .

Posted by: urs on June 17, 2007 11:08 PM | Permalink | Reply to this

Re: Large Smooth Categories

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I suppose the issue is that your vector spaces are not finite dimensional. Otherwise, since any finite dimensional vector space is canonically a manifold and Hom between any two fixed finite dimensional vector spaces is a manifold, the category of vector spaces is internal to the category of finite dimensional manifolds.

On the other hand, you do talk about smooth vector bundles, so your vector spaces must have some extra structure. Are they Banach, Frechet, just topological?

Posted by: Eugene Lerman on June 12, 2007 3:13 AM | Permalink | Reply to this

Re: Large Smooth Categories

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I suppose the issue is that your vector spaces are not finite dimensional.

Hm, not sure. Actually it seems to me as if that is not the central issue. In any case, I would already be happy to completely understand the situation for finite dimensional vector spaces.

Otherwise, since any finite dimensional vector space is canonically a manifold

The problem is, while the vector space itself is a manifold, what we actually need is a smooth structure on the collection of all vector spaces.

In the case of internalization, the idea would be to equip the entity $Obj (Vect)$ (whatever that is: a set, a class? I hope it’s a set. I am no good at telling sets from classes.) with the structure of a smooth space. Any particular vector space would then be a point in that space.

In the pre-stack language, we would regard the pseudofunctor ${Manifolds}^{op} \to Cat$ which sends each manifold to the category of smooth vector bundles over it. The idea is that here, again, we think of a smooth vector bundle as a smooth family of vector spaces.

For that to make sense in the first place it is necessary that any particular vector space itself is a smooth manifold (namely it is a smooth vector bundle over a point). But the subtlety rests in the fact that we need smooth families.

Well, for prestacks here there is not even any real subtlety, I guess. This prestack of smooth vector bundles simply exists.

What I would like to understand is if I have any chance of instead looking at the same entity equivalently as a category internal to smooth spaces.

That might involve set-theoretic issues. But I am not sure I even understand the problem there.

Posted by: urs on June 12, 2007 9:48 AM | Permalink | Reply to this

Re: Large Smooth Categories

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$Obj (Vect)$ is a class. Every set gives rise to a vector space by taking its members as a basis. (Or if you want $Vect$ to be just finite-dimensional vector spaces, then every set gives rise to a one-dimensional space with itself as basis).

Of course, we can get a skeletal version of the category of finite-dimensional vector spaces (or spaces with dimension no more than a fixed cardinal, I think) and its objects will form a set (of cardinality $ℵ_{0}$ in the finite case).

Posted by: Tim Silverman on June 12, 2007 7:30 PM | Permalink | Reply to this

Re: Large Smooth Categories

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$Obj (Vect)$ is a class. Every set gives rise to a vector space by taking its members as a basis. (Or if you want Vect to be just finite-dimensional vector spaces, then every set gives rise to a one-dimensional space with itself as basis).

Of course, we can get a skeletal version of the category of finite-dimensional vector spaces (or spaces with dimension no more than a fixed cardinal, I think) and its objects will form a set (of cardinality ${Aleph}_{0}$ in the finite case).

Okay, thanks. For the application I need it would already be sufficient to restrict to the category ${Vect}_{n}$ of $n$ -dimensional vector spaces.

Suppose I do that, then I am stilled faced with $Mor ({Vect}_{n})$ . Is that a set or a class?

Either way, fixing some manifold $U$ , is the collection of vector bundles over $U$ a set or a class?

And do I really need to care? :-)

All I want to do is get a set of maps $U \to Mor ({Vect}_{n}) .$

Though my problem is that I don’t even see what goes wrong with putting a Chen-smooth structure on ${Vect}_{n}$ if this latter entity turns out to be a class and not a set.

Posted by: urs on June 12, 2007 9:29 PM | Permalink | Reply to this

Re: Large Smooth Categories

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Urs wrote:

That might involve set-theoretic issues. But I am not sure I even understand the problem there.

My undergraduate students have the right attitude here: it doesn’t pay to worry too much about classes!

More precisely: you need to worry about them very hard for a while, until you realize how to avoid worrying about them. Study set theory for a while, ponder Grothendieck’s axiom of universes, and then “set-theoretic issues” will rarely be a problem.

Let’s start by answering your questions in the context of traditional Zermelo-Fraenkel set theory with the Axiom of Choice. In this context, there is no set of all sets. There is no set of 5-element sets, there is no set of all groups, there is no set of all vector spaces, there is no set of 0-dimensional vector spaces, etc.. For any of these things, there is just a proper class. There’s a set of empty sets — there’s just one of those! But for all the rest of the things I listed, there are just ‘too many’ to form a set.

One can work with proper classes: it doesn’t cause instant death to think about them. It’s easier to do this in von Neumann-Bernays-Gödel set theory than Zermelo-Fraenkel set theory. But, category theorists know how to add an extra axiom to ZFC that usually allows us to avoid worrying about proper classes. This is the axiom of universes.

A set $U$ is called a ‘Grothendieck universe’ if it’s closed under a certain list of operations which include most of the things you’re ever likely to do.

Working with sets inside this ‘universe’ is a lot like working with all sets, unless you carry out insanely huge constructions — constructions so monstrous that they burst out of the universe! As a physicist, you’re unlikely to accidentally do something like this.

Sets that are elements of $U$ are often called ‘small sets’. Sets outside $U$ are called ‘large sets’. Speaking metaphorically: from the eyes of someone who lives inside $U$ , large sets seem like proper classes. But, they’re really just sets.

So, for example, the ‘set of all sets in $U$ ’ is a large set… but still a set! It’s just $U$ .

In case one universe is not enough for you, no need to worry! The axiom of universes implies we have nested universes $U \subset U^{'} \subset U ″ \subset \dots$ .

We often call the sets in $U$ small sets, the sets in $U^{'}$ large sets, the sets in $U ″$ extra-large sets (or something silly like that), and so on.

The set of all small sets ( $U$ ) is a large set. The set of all large sets ( $U^{'}$ ) is an extra-large set. And so on…

A category is said to be small (resp. large, extra-large, etc.) if its class of objects and its class of morphisms are small sets (resp. large sets, extra-large sets, etc.).

In this framework, we define $Set$ to be the category of small sets. Thanks to how our first universe $U$ is closed under lots of operations, any sane mathematical physicist should be perfectly happy to work with this version of $Set$ !

So, why bother with the large sets?

Well, note that $Set$ is a large category, since its class of objects is $U$ , a large set — and its class of morphisms is also a large set (not quite so obvious).

So, if we want a version of $Cat$ which includes $Set$ as an object, we had better define $Cat$ to be the category of large categories.

This means that $Cat$ is an extra-large 2-category — it has a large set of objects, a large set of morphisms and a large set of 2-morphisms.

And so on: to get $(n - 1) Cat$ to be an object of $n Cat$ , we need $n Cat$ to be one step larger than $(n - 1) Cat$ .

But: all these large, extra-large, extra-extra-large etc. sets are still just sets — so we never need to worry about proper classes!

In particular, when I talk about $Vect$ , I implicitly mean the category of all small vector spaces. So, like $Set$ , $Vect$ is a large category. So, it’s another object in $Cat$ .

The punchline:

Since $Vect$ has just a (large) set of objects and a (large) set of morphisms, you don’t need to worry about proper classes when pondering the issue of putting a smooth structure on $Vect$ ! Just relax and go ahead. There are certainly interesting puzzles about how to put a smooth structure on $Vect$ , or whether we should bother doing it. But, these questions are very unlikely to involve us in proper classes.

There’s more to say about all this, but until you write a paper about it, I urge you to relax and focus on more serious issues.

Posted by: John Baez on June 13, 2007 9:09 AM | Permalink | Reply to this

Re: Large Smooth Categories

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Let’s start by answering your questions […]

Very helpful, indeed. Thanks a lot for these explanations!

My undergraduate students have the right attitude here: it doesn’t pay to worry too much about classes!

[…]

I urge you to relax and focus on more serious issues.

I’d love to follow that advice. And I am about to. But maybe you could help me by commenting on the following motivation for all these troubles:

While we are busy developing a theory of smooth $n$ -functors by using internalization into a category of smooth spaces, Stephan Stolz and Peter Teichner are in parallel busy doing something very similar. There is more to their work than meets the web, but a glimpse of what they are up to can be found in section 3 of these preliminary notes.

The point is, instead of considering categories internal to sheaves over manifolds (as we have mostly been doing) they consider categories fibred over manifolds, or, equivalently, pseudofuctors from manifolds to categories.

So the natural question is: how are these two approaches related? I started thinking about that in week 20 and later Bruce revived the discussion in week 26 of your “Quantization and Cohomology” lecture.

Personally, I don’t need a smooth structure on $Vect$ to achive happiness, since I think I know better means (very closely related to anafunctors, as you know) to do what this allows us to do. But still, I would like know if, in principle, I could go ahead and put smooth structures on categories like $Vect$ and $n Cob$ by equipping them with Chen-smooth structure (hence realizing them inside sheaves over manifolds) – and how that would compare to instead working over smooth manifolds, as Stolz and Teichner do.

After a lot of back and forth between Bruce and me, I think we were finally able to extract this essence of the question:

how are categories internal to sheaves over some site $S$ related to categories fibred over $S$ ?

Our worry was that categories fibred over $S$ might be inherently more general than those internal to sheaves over $S$ , due to set-theoretic issues.

I think you are saying: no, set-theoretic issues don’t make categories internal to sheaves over $S$ more restrictive than categories fibred over $S$ .

That’s good to hear.

Posted by: urs on June 13, 2007 10:57 AM | Permalink | Reply to this

Re: Large Smooth Categories

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Urs wrote:

Personally, I don’t need a smooth structure on $Vect$ to achieve happiness…

Good! I’m glad it’s not a necessary condition. But, I hope it’s a sufficient condition, because I’d like to make you happy. Let me give it a try.

But still, I would like know if, in principle, I could go ahead and put smooth structures on categories like $Vect$ and $n Cob$ by equipping them with Chen-smooth structure (hence realizing them inside sheaves over manifolds)…

Okay, this is a fun question.

Just to keep things simple and very specific, let’s look at

$Vect = [finite dimensional real vector spaces, linear maps]$

This is a large category, but we don’t let that scare us: we just say it has a large set of objects.

How should we make $Vect$ into a smooth category?

For starters, what should we use as the smooth space of objects? I think we should put the ‘discrete’ smooth structure on the set of vector spaces. This is the one where the only plots are constant maps.

Then, what should we use as the smooth space of morphisms? First, for any pair of objects $V, W \in Vect$ let’s give $hom (V, W)$ its usual smooth structure — it’s a finite-dimensional real vector space. Then, take the disjoint union (i.e., coproduct) of all these smooth spaces $hom (V, W)$ .

It’s easy to check that these choices make $Vect$ into a smooth category.

But is this the ‘correct’ way to make $Vect$ into a smooth category? I think so, because I believe the following 2 things are true. But, it’s late, and I’m tired, so maybe you should check me.

1. For any smooth manifold $X$ (or even any smooth space), let $Disc (X)$ be the smooth category with $X$ as its smooth space of objects, and only identity morphisms. Then a vector bundle over $X$ is ‘the same’ as a smooth anafunctor $F : Disc (X) \to Vect$

More precisely:

Let

$hom (Disc (X), Vect)$

be the groupoid with smooth anafunctors

$f : Disc (X) \to Vect$

as objects and smooth ananatural isomorphisms as morphisms. Then we have an equivalence of groupoids:

$hom (Disc (X), Vect) ≃ [smooth vector bundles over X, smooth vector bundle isomorphisms]$

2. Even better, for any smooth manifold $X$ (or even any smooth space), let $P X$ be the path groupoid of $X$ . Then a vector bundle with connection over $X$ is ‘the same’ as a smooth anafunctor

$F : P X \to Vect$

More precisely:

Let

$hom (P X, Vect)$

be the groupoid with smooth anafunctors

$f : P X \to Vect$

as objects and smooth ananatural isomorphisms as morphisms. Then we have an equivalence of groupoids:

$hom (P X, Vect) ≃ [smooth vector bundles with connection over X,$ $smooth connection - preserving vector bundle isomorphisms]$

Posted by: John Baez on June 16, 2007 10:00 AM | Permalink | Reply to this

Re: Large Smooth Categories

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I’m going to take a risk here and disagree with the last part of John’s comments. But I certainly appreciate his nice explanation of small and large sets, etc.

There are certainly interesting puzzles about how to put a smooth structure on Vect, or whether we should bother doing it. But, these questions are very unlikely to involve us in proper classes.

Basically it is my philosophy that when set-theoretic issues crop up, they are a signal that we should be solving the problem a different way. In other words, often it’s not the actual set-theoretic problem that’s the problem… the real problem is hiding somewhere beneath the surface.

Thus, while I accept that one could use the technology of Grothendieck universes and small and large sets to get around the problem of “vector spaces don’t form a set”, I feel that this solution fails to notice the elephant in the corner, so to speak.

Let me try to identify this elephant in the corner, by offering two concrete objections to Urs’ proposed method to make Vect into a category internal to smooth spaces.

It may be that upon closer examination, the elephant turns out to be just the piled-up remains of last night’s pizza boxes! Time will tell.

Recall Urs’s candidate proposal : we’re going to make Mor(Vect) into a smooth space, by assigning to a manifold $U$ the set of all “smooth families of linear maps parametrized by $U$ ”, i.e.

(1)

Mor (Vect) (U) = {families of linear maps {f_{u}, u \in U}, such that (*) .}

Here $(*)$ means that the family of linear maps $f_{u}$ parametrized by $u \in U$ are the components of a map of vector bundles $F : E \to E^{'}$ over $U$ . Let’s assume set-theoretic qualms about proper classes etc. can be safely ignored.

Objection 1 : Mor(Vect) is not a functor ${Man}^{op} \to Set$ .

Why not? Because it will apply the pull-back construction. A map $g : U_{1} \to U_{2}$ in Man will get sent to the appropriate map of sets

(2)

g^{*} : Mor (Vect) (U_{2}) \to Mor (Vect) (U_{1})

induced by pullback. And pullback isn’t strictly functorial.

Objection 2 : Even if it were a functor, Mor(Vect) wouldn’t be a sheaf.

To be a sheaf would mean that for any covering family $U_{i} \to U$ and every family of elements $f_{i} \in Mor (Vect) (U_{i})$ such that $f_{i} ∣_{ij} = f_{j} ∣_{ij}$ , there would exist a unique $f \in Mor (Vect) (U)$ such that $f ∣_{i} = f_{i}$ .

I claim that such an $f$ doesn’t exist. It does exist if we’re willing to change the equals sign to an isomorphism… but we’re not allowed to do that in the world of sheaves. That’s why stacks were invented!

Given such a covering family, in other words a collection of maps of vector bundles $F_{i} : E_{i} \to E_{i}^{'}$ over each $U_{i}$ , the standard solution would be to glue these together into a map of vector bundles over $U$ by setting, eg.

(3)

E : = \underset{i}{∐} E_{i} / \sim

It’s the disjoint union $∐$ that causes the problem. In the standard setup, the disjoint union of two sets

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June 8, 2007