## June 8, 2007

### Large Smooth Categories

#### Posted by Urs Schreiber

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.

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 $\mathbf{X}$ such that for each manifold $U$ the set $\mathbf{X}(U)$ 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 $\mathbf{X}(U)$ are called plots of $X$.

A puzzling example: Internal smooth structure on $\mathrm{Vect}$?

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 $\mathbf{Mor(Vect)}(U)$ is that of maps $U \to \mathrm{Mor}(\mathrm{Vect})$ which arise as component maps of smooth morphisms $\array{ V_1 &&\stackrel{\phi}{\to}&& V_2 \\ &\searrow && \swarrow \\ && U }$ 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: $\mathrm{Vect}$ as a smooth stack.

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

This should be the same, dually, as a fibred category: thinking of $\mathbf{Vect}$ 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 $\mathbf{Vect} \to \mathrm{Manifolds} \,,$ which is a fibred category.

Questions.

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

How do I decide if $\mathrm{Mor}(C)$ 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 $\mathrm{Mor}(\mathrm{Vect})$?

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

I am confused.

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

$\mathrm{Vect}(\mathrm{point})$ doesn’t look like a groupoid.

I was thinking of stacks with values in $\mathrm{Cat}$ instead of just in $\mathrm{Grpd}$, 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 $\mathrm{Vect}^n$, 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 $\mathbf{Vect} \to \mathrm{manifolds}$ ?

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

### Re: Large Smooth Categories

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 $\mathrm{manifolds} \to \mathrm{Cat}$).

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

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)$\pi : X \rightarrow F$

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

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 $\mathrm{Vect}$ 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

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

$Man \hookrightarrow Shv(Man)$

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 $\{X_\alpha \to X\}$ such that $\cup_\alpha X_\alpha \to X$ is onto.

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

### Re: Large Smooth Categories

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 $X = \mathrm{Mor}(\mathrm{Vect})$? 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

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 $Man^{op} \rightarrow Cat$.

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 $2Cob_s(X)$. It associates to $U \in Man$ the category $2Cob_s(X)(U)$ 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)$Man^op \rightarrow Classes$

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

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

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

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

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

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

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

is a span of ordinary functors $\array{ |F| &\stackrel{f}{\to}& Y \\ \downarrow^p \\ X }$

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 $x \in \mathrm{Obj}(X)$

$f(p^{-1}(\mathrm{Id}_x))$

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

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

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

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

$\mathrm{Hom}_D(-,C)$

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 $\mathrm{DiscCat}_D \subset \mathrm{Cat}_D$ of all those categories internal to $D$ which are discrete in that they only have identity morphisms.

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

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

$X \to \mathrm{our}\; \mathrm{Stack}$

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 $G\mathrm{Tor}$ (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 $G\mathrm{Tor}$ as internal to sheaves over manifolds.

Do we get this from what you said? Probably the point is that while $G$ and $G\mathrm{Tor}$ 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

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

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

$\Sigma G \simeq G\mathrm{Tor}$

(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 $\Sigma G$ as a stack.

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

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

### Re: Large Smooth Categories

ΣGGTor 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

Urs wrote:

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

$\Sigma G \simeq G Tor$

(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, $\Sigma G$ 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

$hom(X,Y) \cong G$

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

$A: G Tor \to \Sigma G$

$B: \Sigma G \to G Tor$

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

$\alpha: B \circ A \Rightarrow 1_{G Tor}$

and

$\beta: A \circ B \Rightarrow 1_{\Sigma G}$

If I’m right, this means that $G Tor$ and $\Sigma G$ 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

$2C^\infty = [smooth categories, smooth anafunctors, smooth ananatural transformations]$

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

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

$hom(P X, \Sigma G) \simeq$ $[smooth principal G -bundles with connection over X, isomorphisms thereof]$

where $hom(P X, \Sigma G)$ is defined in $2C^\infty$. In other words, $hom(P X, \Sigma G)$ is the category with smooth anafunctors

$F : P X \to \Sigma G$

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

If we indeed have

$\Sigma G \simeq G Tor$

in $2 C^\infty$, we can instantly conclude that we also have

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

where $hom(P X, G Tor)$ is again defined in $2 C^\infty$.

If everything I’m saying is true, $G Tor$ and $\Sigma G$ 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

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 $G\mathrm{Tor}$ such that ordinary Chen smooth functors

$P X \to G\mathrm{Tor}$

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

If we use anafunctors, we can just as well stick with $\Sigma G$ as a model for $G\mathrm{Tor}$ 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 $G\mathrm{Tor}$.

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

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

$hom(P X, G Tor) \simeq [smooth principal G−bundles with connection over X,isomorphisms thereof]$

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

$C^\infty Cat = [smooth categories, smooth functors, smooth natural transformations] ?$

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

Hi John,

only have time for a very quick reply:

yes, I expect that smooth functors from paths to $G\mathrm{Tor}$, 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 $G\mathrm{Tor}$ 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 $G\mathrm{Tor}$ 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 $\mathrm{Disc}(X)$ to $G\mathrm{Tor}$. 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 $G\mathrm{Tor}$.

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 $G\mathrm{Tor}$ 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

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

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 $\mathrm{Obj}(\mathrm{Vect})$ (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 $\mathrm{Manifolds}^{\mathrm{op}} \to \mathrm{Cat}$ 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

$Obj(Vect)$ 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).

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

### Re: Large Smooth Categories

$\mathrm{Obj}(\mathrm{Vect})$ 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 $\mathrm{Vect}_n$ of $n$-dimensional vector spaces.

Suppose I do that, then I am stilled faced with $\mathrm{Mor}(\mathrm{Vect}_n)$. 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 $U \to \mathrm{Mor}(\mathrm{Vect}_n) \,.$

Though my problem is that I don’t even see what goes wrong with putting a Chen-smooth structure on $\mathrm{Vect}_n$ 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

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 $U \subset U^' \subset U'' \subset \cdots$.

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

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 $\mathrm{Vect}$ 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 $\mathrm{Vect}$ and $n\mathrm{Cob}$ 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

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:

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

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:

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

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

### Re: Large Smooth Categories

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 $F : E \rightarrow E^'$ over $U$. Let’s assume set-theoretic qualms about proper classes etc. can be safely ignored.

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

Why not? Because it will apply the pull-back construction. A map $g : U_1 \rightarrow U_2$ in Man will get sent to the appropriate map of sets

(2)$g^* : Mor(Vect)(U_2) \rightarrow Mor(Vect)(U_1)$

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 $U_i \rightarrow U$ 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 $F_i : E_i \rightarrow E^'_i$ 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)$E := \coprod_i E_i / \sim$

It’s the disjoint union $\coprod$ that causes the problem. In the standard setup, the disjoint union of two sets $A$ and $B$ will be the set of ordered pairs

(4)$\{ (a, 1), a \in A \} \union \{(b, 2), b \in B \}.$

Applying this to glue together our vector bundle $E$, we see that there’s no way it can restrict to exactly $E_i$… because $E|_i$ has elements which are equivalence classes of ordered pairs, while $E_i$ has just plain elements! Granted, they’re definitely canonically isomorphic… but that’s the whole reason I am suggesting to rewrite this in terms of (pre)stacks or categories fibered over manifolds, where we don’t have this problem.

Let me close by saying that “objection 1” above is just a rewording of Urs’ own misgivings about this setup in his notes on the subject. There he notices the sublety about “existence of a bundle” versus the bundle itself. I’ve written things out in the “concrete sheaf” paradigm of smooth spaces; if you write them out in the “plot” paradigm then one runs into this sublety too.

Posted by: Bruce Bartlett on June 13, 2007 12:16 PM | Permalink | Reply to this

### Re: Large Smooth Categories

I am prepared to recognize an elephant right there in the corner – if it were not for that faint smell of pizza.

Bruce argues above that the assignment

$U \mapsto \left\{ f : U \to \mathrm{Mor}(\mathrm{Vect}) | \exists \array{ E_1 &&\stackrel{f}{\to}&& E_2 \\ & \searrow && \swarrow \\ && U \; } \in \mathrm{Mor}(\mathrm{VectBun}(U)) \right\}$

is not a sheaf, since pullback of vector bundles respects composition only weakly.

He then emphasizes again the subtlety encoded in the $\exists$-quantifier.

But I am wondering: isn’t it precisely that subtlety here which does make the above a sheaf?

The whole point is that instead of assigning sets of vector bundle morphisms to $U$, I want to assign maps from $U$ to somewhere to $U$, such that some appropriate vector bundle morphism exists.

Now, maps do pull back strictly. So the question really is: if $\phi : V \to U$ is given and $f : U \to \mathrm{Mor}(\mathrm{Vect})$ is such that there are two vector bundles $E_1$ and $E_2$ such that $f$ is the component map of a smooth morphism between them, do there exist vector bundles $\tilde E_1, \tilde E_2$ over $V$, such that $f \circ \phi$ is the component map of a morphism between them?

I’d think it is clear that there is: simply take $\tilde E_i := \phi^* E_i$.

So doesn’t that run into a contradiction then, since the isomorphism $\phi_2^* (\phi_1^* E_i) \simeq (\phi_2 \circ\phi_1)^* E_i$ is nontrivial, while $\phi_2^* (\phi_1^* f) = (\phi_1 \circ\phi_2)^* f$ ?

I am inclined to think that there is no contradiction, precisely due to the $\exists$-quantifier!

Because this admits the freedom that there is more than one pair of vector bundles such that our function $U \to \mathrm{Mor}(\mathrm{Vect})$ is the component map of a morphism between them.

Namely we find that $(\phi_1 \circ \phi_2)^* f$ is the component map of bundle morphisms for two different pairs of bundles: on the one hand $\phi_2^* (\phi_1^* E_1) \to \phi_2^* (\phi_1^* E_2)$ and on the other hand $(\phi_1 \circ \phi_2)^* E_1 \to (\phi_1 \circ \phi_2)^* E_2 \,.$

Hm. I am saying this mainly to make somebody make me see why I am all wrong. :-)

Posted by: urs on June 13, 2007 1:33 PM | Permalink | Reply to this

### Re: Large Smooth Categories

Ah, it seems I just showed myself that I need to be more careful with stating that condition on the maps $U \to \mathrm{Mor}(\mathrm{Vect})$. Above I used the word “is”, that’s dangerous.

Judging from the end of my above argument, what I am really thinking of is that $f : U \to \mathrm{Mor}(\mathrm{Vect})$ is a plot if after choosing isomorphisms of the vector spaces involved, it is the component map of a smooth vector bundle morphism.

Then it is this choice of isomorphisms which absorbs the failure of pullback of vector bundles to compose strictly.

Posted by: urs on June 13, 2007 2:07 PM | Permalink | Reply to this

### Re: Large Smooth Categories

Urs is a worthy opponent; I feel that I am the evil criminal trying to quash his resistance in a Die Hard movie. Managed to escape the elevator shaft, eh? Here is the next round.

Since we are prepared to believe that $Mor(Vect)(U)$ can be regarded as a set, we must be explicit about what it means for two elements of this set to be equal… the “criterion of equality”.

Now elements of $Mor(Vect)(U)$ are families of linear maps between vector spaces. I believe you will agree that $f, g \in Mor(Vect)(U)$ are equal if and only if

(a) For each $u \in U$, domain($f(u)$) = domain($g(u)$) and range($f(u)$) = range($g(u)$).

(b) For each $u \in U$, $f(u) : domain(f(u)) \rightarrow range(f(u))$ is equal to the linear map $g(u)$.

It’s condition (a) that makes $Mor(Vect)$ fail to be a sheaf. ( Warning : It is very subtle to see this if you think in terms of plots… though the problem is still there. I encourage you, for this problem, to think of a smooth space as a concrete sheaf.)

So let $f \in Mor(Vect)(U)$. If there exist vector bundles $E$ and $E^'$ above $U$ such that $f(u)$ is the component of a vector bundle map betwen $E$ and $E^'$, then by definition the domains and ranges of the maps $f(u)$ are the fibers of these bundles, i.e.

(1)$Domain(f(u)) = E_u, Range(f(u)) = E^'_u$

And that’s where the problem lies. Because now pullback-not-being-strictly-functorial becomes serious, since we can now distinguish $(\phi_1 \circ \phi_2)^* f$ from $\phi_2^*(\phi_1^* f)$ : these linear maps will have different domains and ranges, as $u$ varies over $U$.

It’s all very subtle and can make your head spin, but there it is.

Posted by: Bruce Bartlett on June 13, 2007 2:10 PM | Permalink | Reply to this

### Re: Large Smooth Categories

Bruce, right, I am thinking while typing. Did manage to evade your latest assault before you even hit submit. ;-)

Yes, I agree with what you just said. But in my evil ways, I am already thinking of other strategies.

Not sure if anybody out there will still think it is a worthy endeavour to try to realize $\mathrm{Vect}$ internal to sheaves over manifolds, but right now I have some momentum – so here I go.

A minute ago I fell in love with this solution:

Assign to each manifold $U$ the collection of natural transformations $\array{ & \nearrow \searrow^{E_1} \\ P_0(U) & \Downarrow^{f}& \mathrm{Vect} \\ & \searrow \nearrow_{E_2} }$ of transport functors. So this are just any old such transformations of functors, such that we may equip everything in sight with certain smoothness conditions.

Notice that this is really nothing but making precise what I said in my latest remark. I think.

The point is, these gadgets do pull back strictly. On objects and on morphisms. And if one of these admits a smooth structure, then its pullback does.

Hah, now that’s an idea! Bruce will have a hard time disabusing me of that one! ;-)

In a word, now I am saying: let’s consider the sheaf $U \mapsto \mathrm{Trans}(P_0(U),\mathrm{Vect})$ which assigns to each manifold $U$ the category of transport functors on constant paths with values in vector spaces.

How can that be, these are supposedly the same as vector bundles?

Well, there is a change of perspective here: instead of considering a vector bundle to be a total space with certain structure and properties, this realizes vector bundles as fiber-assigning-functors.

And these do pull back strictly!

Posted by: urs on June 13, 2007 2:24 PM | Permalink | Reply to this

### Re: Large Smooth Categories

I realize I should state my proposal more cleanly:

There is an alternative to regarding fiber bundles as total spaces $E \to X$ and so on: instead, we may think of them in terms of fiber-assigning-functors

$E : (P_0(X) := \mathrm{Disc}(X)) \to \mathrm{WhateverCategoryOfFibers}$

with a certain property. That property I am calling the property of being a transport functor. The full details of this are described and worked out here.

Transport functors are equivalent to the corresponding fiber bundles. But they are still a little different.

For one, pulling back a transport functor just amounts to precomposing it:

for $\phi : V \to U$ and $\mathrm{E} : \mathrm{Disc}(U) \to \mathrm{Vect}$ a transport functor, we have $\phi^* E : \mathrm{Disc}(V) \stackrel{\phi}{\to} \mathrm{Disc}(U) \stackrel{E}{\to} \mathrm{Vect} \,.$ This clearly respects composition of strictly.

A morphism of transport functors is just an ordinary morphism of functors, with the obvious property that it is compatible with the transport property.

We get a category $\mathrm{Trans}(\mathrm{Disc}(U), \mathrm{Vect})$ this way. With the remaining details filled in, this is equivalent to that of smooth vector bundles over $U$.

So now I am proposing to realize the category $\mathrm{Vect}$ as internal to sheaves over manifolds as follows:

let the sheaf of objects be

$U \mapsto \mathrm{Obj}( \mathrm{Trans}(\mathrm{Disc}(U), \mathrm{Vect}) ) \,;$

let the sheaf of morphisms be

$U \mapsto \mathrm{Mor}( \mathrm{Trans}(\mathrm{Disc}(U), \mathrm{Vect}) ) \,.$

I am thinking that this makes precise what I said previously:

the sheaf of morphisms assigns to $U$ the set of maps $U \to \mathrm{Mor}(\mathrm{Vect})$ which are such that they “come from” component maps of smooth vector bundle morphisms over $U$.

The transport functor formalism makes this “comes from” precise.

Posted by: urs on June 13, 2007 3:48 PM | Permalink | Reply to this

### Re: Large Smooth Categories

For what it’s worth, I enjoy watching you guys go back and forth much more than any Die Hard movie :)

*chomping popcorn*

Posted by: Eric on June 13, 2007 3:18 PM | Permalink | Reply to this

### Re: Large Smooth Categories

I enjoy watching you guys go back and forth much more than any Die Hard movie

There is also a director’s cut version including previously unseen email correspondence. But that’s not suited for all audiences…

Posted by: urs on June 13, 2007 3:23 PM | Permalink | Reply to this

### Re: Large Smooth Categories

I think part B of Sketches of an Elephant is an excellent reference for internal categories in sheaves over $S$ and categories fibered over $S$. Another is A Homotopy Theory for Stacks by Sharon Hollander, arXiv:math/0110247. Here’s a summary of the philosophy, as I see it.

Let $E$ be a category with finite limits, such as a category of sheaves. Then a category internal to $E$ may be thought of as a small category in the world of $E$’ (that is, a small category when we regard $E$ as a replacement for the category of sets), while a category fibered over $E$ may be thought of as a large category in the world of $E$’.

Passing to stacks, or H-S maps, or anafunctors, or inverting weak equivalences’, are ways of making these categories in the world of $E$ behave more like the Set-based categories we are used to. For example, it helps us get around the fact that $E$ doesn’t in general satisfy the internal axiom of choice’.

If its fibers are small, then a category fibered over $E$ is equivalent to a pseudofunctor $E^{\mathit{op}}\to\mathit{Cat}$. Such a pseudofunctor can always be rectified’ into a strict functor, and even made into a sheaf of categories, without changing the stack it represents up to equivalence. In general this is an unnatural procedure, like the replacement of a monoidal category by a strict one, and it’s only true once we’ve passed to stacks.

Now, a sheaf of small categories on $E$ is the same as an internal category in sheaves on $E$, and if $E$ is itself the category of sheaves on some small site, then every sheaf on $E$ is representable. Thus every small stack on such an $E$ is represented by an internal category in $E$.

A stack over $E$ whose fibers are large may still be made into a sheaf of large categories, which is thus equivalent to an internal category in the category of large sheaves’ on $E$, or the site defining it. If we take the point of view that all we care about is the site, then we may feel free to replace $E$ by the category of large sheaves on the same site and work that way.

However, I’m not convinced that this is always the best thing to do. For instance, it completely avoids the question of what the analogue is in the world of $E$ of a locally small category, that is, a large category whose hom-sets are small. There is, however, a good notion of locally small $E$-category’ for fibered categories. And as unpalatable as locally small categories may be for some formal manipulations, the fact is that almost all large categories in nature are locally small, and local smallness is a very useful property. For example, you can take a coproduct over a hom-set in a locally small category that has small colimits. Moreover, if you want to generalize to enriched category theory, then locally small categories are the right thing to look at, since they are the same as Set-enriched categories.

Another argument against just using large sheaves is that it only works when $E$ is already a Grothendieck topos (the category of small sheaves on some small site). This is certainly an interesting and important case, but it seems to me that it is far from the only situation in which we would want to talk about internal categories and stacks. Even the category of Chen-smooth spaces is not a Grothendieck topos. I believe it is a concrete quasitopos’, but that’s not good enough to make any sheaf on it representable. (Note: the quasi’ in quasitopos’ has absolutely nothing to do with the quasi’ in `quasicategory’.)

Thus, I believe that while invoking Grothendieck universes and working with large sheaves may work in some cases, it is in general a misleading avenue. I’m not sure I agree with Bruce that set-theoretic issues are always a signal that we should be solving the problem in a different way, but here at least I do think that is the case.

Posted by: Mike Shulman on June 18, 2007 5:25 PM | Permalink | Reply to this

### Re: Large Smooth Categories

If its fibers are small, then a category fibered over $E$ is equivalent to a pseudofunctor $E^{\mathrm{op}} \to \mathrm{Cat}$.

Such a pseudofunctor can always be ‘rectified’ into a strict functor, and even made into a sheaf of categories, without changing the stack it represents up to equivalence.

Ah, that’s what I was looking for. Is this discussed in Hollander’s text? (I don’t have “Sketches of an Elephant” available at the moment.)

In general this is an unnatural procedure, like the replacement of a monoidal category by a strict one,

I can believe that.

A stack over $E$ whose fibers are large may still be made into a sheaf of large categories, which is thus equivalent to an internal category

[…]

However, I’m not convinced that this is always the best thing to do.

[…]

I’m not sure I agree with Bruce that set-theoretic issues are always a signal that we should be solving the problem in a different way, but here at least I do think that is the case.

Very useful, thanks! So in summary: Bruce is morally right to favor the stacky picture, while it is still true that – by using force – one may remain in the internal picture if desired.

Posted by: urs on June 18, 2007 7:59 PM | Permalink | Reply to this

### Re: Large Smooth Categories

I would like to emphasize that whatever the outcome of this discussion is, I am glad it’s taking place and I think I am learning quite a bit. I am indebted to Bruce for his insistence.

With a little luck, he will have managed to convince me of his point of view soon. :-)

Posted by: urs on June 13, 2007 1:39 PM | Permalink | Reply to this

### Re: Large Smooth Categories

I apologize if this is a bit off-topic.

I am wondering what’s known about parallel transport in stacks. It seems that there is a good notion of a connection on a principal bundle over a Deligne-Mumford stack (for general stacks there seems to be a problem; see Behrend-Xu paper mentioned earlier). So there should be a notion of parallel transport. Has this been spelled out anywhere?

Posted by: Eugene Lerman on June 13, 2007 5:43 PM | Permalink | Reply to this

### Re: Large Smooth Categories

It seems that there is a good notion of a connection on a principal bundle over a Deligne-Mumford stack

Can you briefly indicate what kind of deifnition you have in mind – the main idea?

Personally, I am really interested in connections only in as far as they admit parallel transport. So my first question would be:

given a stack $\mathbf{X}$, sufficiently well behaved, what would be the analogue of the groupoid of thin-homotopy classes of path in that?

Objects should be the points of $\mathbf{X}$, morphisms should be classes of morphisms $\mathbf{[0,1]} \to \mathbf{X} \,,$ where now the standard interval is identitfied with a stack in the usual way $\mathbf{[0,1]} : U \mapsto C^\infty(U,[0,1]) \,.$

The two technical questions to be solved are:

can we give a stacky formulation of the two conditions which make composition of such morphisms well defined:

1) we want to say that we require these maps to be constant near a neighbourhood of the boundary of $[0,1]$

2) we want to identify two such maps if they are connected by a homotopy whose differential has everywhere non-maximal rank.

I guess the first condition is easy to implement. The second might be more involved. But maybe if we follow David Roberts’ suggestion to reformulate this in terms of trees, then it’s easier.

Anyway, once any such notion of smooth path groupoid of a stack exists, there is an obvious way to define a connection with parallel transport in that context.

Posted by: urs on June 13, 2007 6:04 PM | Permalink | Reply to this

### Re: Large Smooth Categories

A principal G bundle over an Artin stack X is a functor from X to the stack BG (the objects of BG are principal bundles, the morphisms are equivariant maps of bundles). If X_\bullet is a groupoid representing the stack X, then a map from X_\dot to BG is a X_\bullet - G bibundle, which is a principal G-bundle over X_0 with the action of X_\bullet.

Now let’s add connections to this picture. Replace BG by the following stack: objects are principal G-bundles with connections (ie. Lie(G)-valued 1-forms), arrows are equivariant maps preserving the connections. Let’s call this stack DBG (perhaps one can come up with a better name) Then a connection on X is a functor to DBG. In terms of groupoids this amounts to a principal G-bundle P_0 over the space of objects X_0 which is X_\bullet invariant and a connection 1-form, which is X_\bullet invariant and X_bullet basic.

Either one of the two descriptions is, for me, a principal bundle with a connection over my stack X.

When we start talking about a path groupoid of a stack, things get a little more interesting. The collection points in a stack is not just a set, it’s a groupoid. The “set” of paths between two fixed points in a stack is not a “set”, it’s again a groupoid. So it looks like the fundamental groupoid of a stack is a groupoid in groupoids. Hmm…

Posted by: Eugene Lerman on June 13, 2007 8:27 PM | Permalink | Reply to this

### Re: Large Smooth Categories

Eugene,

thanks! I’ll reply to that in stages. Here is stage number one:

A principal $G$-bundle over an Artin stack $X$ is a functor from $X$ to the stack $B G$(the objects of $B G$ are principal bundles, the morphisms are equivariant maps of bundles).

Just for the record, let me translate this into the nomenclature that Bruce has been using, since that might help make the different pictures here converge a little more.

Bruce suggested to call the stack of $G$-bundles $G\mathrm{Tor}_s$ because to a point it assigns the category (a groupoid) we like to call $G\mathrm{Tor}$: objects are $G$-spaces which are isomorphic to $G$ as $G$-spaces, morphisms are the obvious morphisms of these.

Sticking to the idea of describing a $G$-bundle over a space $X$ by its “fiber-assigning functor” $P : \mathrm{Disc}(X) \to G\mathrm{Tor}$ which sends every point of $X$ to the fiber living over it (which is a $G$-torsor, an object in $G\mathrm{Tor}$), we need to equip this functor with the required smoothness property.

One way to do this, as Bruce emphasized so much, is to promote everything in sight to stacks (“passing to the families version”).

Hence a smooth $G$-bundle over an ordinary smooth space $X$ is a morphism of stacks $P_s : \mathrm{Disc}(X)_s \to G\mathrm{Tor}_s \,.$ Here I am writing $\mathrm{Disc}(X)_s$ for the ordinary stack associated with any space $X$: $\mathrm{Disc}(X)_s : U \mapsto C^\infty(U,X)$ to emphasize how this is the families version of a functor $\mathrm{Disc}(X) \to G\mathrm{Tor}$.

One reason why I think it is worth emphasizing this is the following:

With that in place, we can then replace the stack $\mathrm{Disc}(X)_s$ with any other (suitably well behaved) stack $\mathbf{X}$ and say that a morphism of stacks $P : \mathbf{X} \to G\mathrm{Tor}_s$ is a $G$-bundle over $X$.

In many cases $\mathbf{X}$ will encode some kind of smooth groupoid to be thought of as the corresponding orbifold. With $\mathbf{X}(\mathrm{pt})$ that groupoid, the morphism $P$ of stacks gives us in particular a functor $P(\mathrm{pt}) : \mathbf{X}(\mathrm{pt}) \to G\mathrm{Tor} \,.$ This is indeed an equivariant $G$-bundle on $\mathrm{Obj}(\mathbf{X}(\mathrm{pt}))$, as it should be for a bundle on an orbifold.

Posted by: urs on June 13, 2007 8:49 PM | Permalink | Reply to this

### Re: Large Smooth Categories

Second stage.

Now let’s add connections to this picture. Replace $B G$ by the following stack: objects are principal $G$-bundles with connections

Interestingly, at this point the setup deviates from what we have been talking about so far. I’d be interested in better understanding the relation between the two.

In all the messages with which Bruce and myself have recently cluttered up the $n$-Café – and of course in everything that John has been talking about in the Quantization and Cohomology lecture – the premise was that a $G$-bundle with connection on a space $X$ would be a functor $\mathrm{tra} : P_1(X) \to G\mathrm{Tor}$ which smooth in some suitable sense. Here $P_1(X)$ denotes the groupoid of paths in $X$.

If we follow the families way of thinking using stacks (or at least prestacks) we’d say, as Bruce has so much emphasized here, that this functor is smooth if it happens to be the value over a point of a morphism of stacks $\mathrm{tra}_s : (P_1(X))_s \to G\mathrm{Tor}_s \,,$ where now $(P_1(X))_s$ is the stack which assigns to any manifold $U$ the groupoid of smooth $U$-families of points in $X$ and smooth $U$-families of paths in $X$.

What you describe seems to be different in spirit from this line of thinking. Whereas here the parallel transport, in a way, is encoded in the domain stack (since that knows about paths in $X$), in what you describe the domain is the same as for bundles without connection, and the connection enters through the codomain stack, which is now taken to be that of $G$-bundles with connection.

This is something I need to think about.

Posted by: urs on June 13, 2007 9:11 PM | Permalink | Reply to this

### Re: Large Smooth Categories

Urs,

Let me add something I should have emphasized in my previous post: there are principal bundles over stacks that don’t have connection 1-forms. This is something I learned from the paper of Behrend and Xu mentioned earlier. And this is probably the reason why they use the so called “pseudo-connections” to define characteristic classes.

In a way the non-existence of connection 1-forms on Artin stacks is not surprising: Artin stacks, in general, don’t have tangent/cotangent bundles. But for Deligne-Mumford stacks a principal $G$-bundle with a connection 1-form as a map from the stack into $DBG$ works quite well.

And to clear a possible confusion: a stack with an atlas is what one usually calls an Artin stack.

Posted by: Eugene Lerman on June 14, 2007 2:30 AM | Permalink | Reply to this

### Re: Large Smooth Categories

Eugene, I don’t know if this is relevant to the discussion, but I believe one can define connections on an Artin stack in a reasonable way by replacing the tangent bundle by the tangent complex (with its $L_\infty$ structure), so that presumably connection one-forms live in the cotangent complex (or at least your colleague Tom Nevins convinced me of something like this!). Certainly for a smooth Artin stack the corresponding notion of flat connection is the “correct” one - i.e. it agrees with several other reasonable definitions, such as defining flat connections by descent from a smooth cover, or bundles with infinitesimal parallel transport to all orders (equivariance for the de Rham groupoid). For singular spaces (not to mention stacks) there are several competing notions of flat bundle, and the “morally correct” one (crystals) doesn’t have much to do with tangent bundles, at least naively. Hope this helps, David

Posted by: David Ben-Zvi on June 14, 2007 4:10 AM | Permalink | Reply to this

### Re: Large Smooth Categories

there are several competing notions of flat bundle, and the “morally correct” one (crystals) doesn’t have much to do with tangent bundles

Could you briefly say what the idea behind the definition of a “crystal” is, in this context? Is there a good reference where to read about it?

Posted by: urs on June 14, 2007 12:11 PM | Permalink | Reply to this

### Re: Large Smooth Categories

Urs, A crystal is an object with infinitesimal parallel transport, or to quote Grothendieck, it is something which grows and is rigid. Let me restrict to characteristic zero, so we can ignore subtleties with divided powers. Then there are two main variants on the idea. For references (outside of the big crystalline cohomology literature) I’d recommend Simpson’s writings on nonabelian Hodge theory, where he explains nonabelian versions of the Gauss-Manin connection in this context. Teleman’s Inventiones paper on Borel-Weil-Bott has an appendix describing some of this and he and Simpson have a beautiful unpublished paper (on Constantin’s webpage) on de Rham’s theorem for infinity stacks, where these ideas are pushed very far..

First we have the de Rham groupoid (really an equivalence relation) on any scheme X which is given by the formal neighborhood of the diagonal in X times X. This is the graph of the equivalence relation of infinitesimal nearness. For X smooth a vector bundle with flat connection is the same as a vector bundle equivariant for this groupoid. Now you can replace vector bundle by space over X, or sheaf of categories, or sheaf of infinty-categories, etc and you get a notion of flatness for any of these. This just encodes parallel transport to any finite order. This is Grothendieck’s notion of a stratification.

When X is singular, this notion is a little too naive. A crystal (of vector spaces, or spaces, or categories, or….) is an object equipped with parallel transport for all infinitesimal thickenings — actually this notion probably works very well in non algebro geometric contexts: replace infinitesimal thickenings by some kind of homotopy retracts). namely if U in X has an infinitesimal thickening V, then we are given an identification of our sheaf on U and on V, and this needs to be done compatibly for all V. i.e. whenever you thicken U in such a way that the topology doesn’t change, your crystal should grow rigidly from U to V – what you do on V must be determined by what you do on U.

This gives a great definition of flat objects of any kind on any space. An example: the category of local systems on a space is a crystal of categories. Namely local systems on U and V are just plain the same. In particular if you vary a space X, any local system can be varied canonically with X, in such a way that it extends to a local system on the total space of the family.. this is known as isomonodromic deformation.

Posted by: David Ben-Zvi on June 18, 2007 10:36 PM | Permalink | Reply to this

### Re: Large Smooth Categories

Simpson s writings on nonabelian Hodge theory, where he explains nonabelian versions of the Gauss-Manin connection

Thanks. Is that at all similar to the work by Breen and Messing on nonabelian differential cohomology?

(I am just asking because Breen and Messing I am resonably familiar with, and it seems that the tools you are mentioning are essentially those also used by them.)

A crystal (of vector spaces, or spaces, or categories, or…)
is an object equipped with parallel transport for all infinitesimal thickenings

Just so that I can be sure I know what is going on: would you mind giving me the precise definition of “infinitesimal thickening”?

Posted by: urs on June 18, 2007 11:07 PM | Permalink | Reply to this

### Re: Large Smooth Categories

Urs,

By infinitesimal thickening V of U I mean U is closed in V and defined by a nilpotent (sheaf of) ideals. You can use such thickenings V of opens U to define a Grothendieck topology (the infinitesimal site, I believe) – see Grothendieck’s Crystals and the de Rham cohomology of schemes, or work of Berthelot. Anyway I think if you use instead of this any kind of thickening which is topologically trivial you should get a good notion.

When X is smooth any such thickening V of U open in X splits back onto U. As a result if you have a sheaf equivariant for the formal nbhd of the diagonal it already knows all of this infinitesimal thickening info, so stratifications = crystals.

I don’t know the Breen-Messing stuff well, but what I’m talking about is all notions of integrable/flat connections, so orthogonal to their discussion of curvature. If you’re on a smooth space you can use the first order nbhd of the diagonal to define connections (rather than flat connections), on a singular space you would need the tangent complex..

And now a message from our sponsors: I cannot resist giving my default reference for many questions, Beilinson-Drinfeld’s tome on Quantization of Hitchin’s hamiltonians. Its legendary (at least in some circles!) Chapter 7 is a standalone book on D-modules (eg flat connections) on stacks, Hecke operators and other ideas, explained in a demanding but inimitable way. In particular notions of flat connections on stacks are addressed from at least three points of view in there (including the crystalline one), and a very general Hecke story (probably relevant to the ongoing Tale of Groupoidification) is told…

Posted by: David Ben-Zvi on June 18, 2007 11:30 PM | Permalink | Reply to this

### Re: Large Smooth Categories

Artin stacks, in general, don’t have tangent/cotangent bundles.

The way I think about it is this: a stack $X$ (over manifolds, say) is a way to equip a groupoid with a kind of smooth structure, namely the groupoid $X(\mathrm{point})$.

If I want to think of stacks as generalized spaces (as opposed to conceiving them as smooth groupoids, which are 2-spaces) I am thinking of this groupoid as encoding an orbifold:

the morphism in the groupoid connect precisely those points which are connected by the group action defining the orbifold.

Now, what would, then, be the tangent bundle of such a beast?

I find it conceptually helpful here to think in terms of integrated structures:

the tangent bundle of a smooth space $X$ I may think of as the groupoid of thin-homotopy classes of paths $P_1(X)$ in $X$: that’s because its dual $\mathrm{Hom}(P_1(X), \Sigma \mathbb{R})$ is precisely the space of 1-forms on $X$.

(Above and further on now, everything in sight is in the context of smooth spaces.)

We should be thinking now of the space $X$ as the groupoid $\mathrm{Disc}(X)$, which has only identity morphisms.

Then, what would be the analog of $P_1(X)$ as we pass from $\mathrm{Disc}(X)$ to some more general groupoid $\mathrm{Gr}$?

Suppose, for example, that

$\mathrm{Gr} = Y \times_X Y \stackrel{\stackrel{\pi_1}{\to}}{ \stackrel{\to}{\pi_2}} Y$

is the groupoid induced by a cover

$Y \to X \,.$

Then the right groupoid which encodes the “integrated tangent space” of $\mathrm{Gr}$ is the weak pushout of

$\array{ P_1(Y \times_X Y) &\to& P_1(Y) \\ \downarrow \\ P_1(Y) } \,.$

This weak pushout groupoid – I am now used to calling it $C_\bullet(Y)$ – has a very natural interpretation:

it is generated from paths in the cover $Y$ together with “jumps” $\pi_1(y) \to \pi_2(y)$ between the covering pathches, modulo the relation

$\array{ \pi_1(x) &\stackrel{\pi_1(\gamma)}{\to}& \pi_1(x) \\ \downarrow && \downarrow \\ \pi_2(x) &\stackrel{\pi_2(\gamma)}{\to}& \pi_2(x) }$ for all paths $x \stackrel{\gamma}{\to} y$ in $Y \times_X Y$.

Of course, just as $Y \times_X Y$ is just a blown-up version of $X$, this is just a blown-up version of $P_1(X)$.

But it serves as an indication for what goes wrong with “tangent bundles” for more general groupoids:

for $\mathrm{Gr}$ any smooth groupoid, we would expect its “integrated tangent bundle” (its path groupoid) to be that groupoid which is generated from paths in $\mathrm{Obj}(\mathrm{Gr})$ together with morphisms, i.e. points in $\mathrm{Mor}(\mathrm{Gr})$ – modulo some relations.

For general groupoids $\mathrm{Gr}$, it is not clear (to me at least) what these relations would be. The relation used above rested on the fact that $Y \times_X Y$ is a poset (at most one morphism for every ordered pair of points).

On the other hand, this way of thinking about tangent spaces of groupoids might be one we to understand why, as David Ben-Zvi mentioned it is still possible to define at least flat connections:

for flat connections, we don’t need the full path groupoid $P_1(X)$, but just the fundamental groupoid $\Pi_1(X)$. Now this is locally codiscrete! And when we want to differentiate, only the local structure counts.

This might be what underlies the existence of good definitions of flat connections on stacks, I could imagine.

Posted by: urs on June 14, 2007 11:14 AM | Permalink | Reply to this

### Re: Large Smooth Categories

Artin stacks, in general, don’t have tangent/cotangent bundles.

Here is yet another way to maybe get some intuition for that:

a smooth stack, “being” a smooth groupoid, should not have a tangent bundle, but a tangent 2-bundle.

An easy way to see this is to instead consider the somewhet simpler situation of a groupoid $\mathrm{Gr}$ internal to smooth spaces.

So, that’s in particular a diagram

$\mathrm{Mor}(\mathrm{Gr}) \stackrel{s,t}{\to} \mathrm{Obj}(\mathrm{Gr})$

of smooth spaces $S^\infty$. We may then hit everyhting in sight with the tangent functor

$T : S^\infty \to \mathrm{VectBun}$

to get a category internal to that of vector bundles:

$T \mathrm{Mor}(\mathrm{Gr}) \stackrel{s_*,t_*}{\to} T \mathrm{Obj}(\mathrm{Gr})$

That’s one way to think of the “tangent 2-bundle”. Danny Stevenson is using that to great effect in his description of connections on 2-bundles (“nonabelian gerbes”).

Posted by: urs on June 14, 2007 12:22 PM | Permalink | Reply to this

### Re: Large Smooth Categories

Hi Urs,

This is likely not the best place to ask such a terribly basic question, but your response made me think of it, so this may be as good a place as any…

I won’t pretend to understand “stacks”, much less tangent 2-bundles. I struggle with even simple tangent bundles.

I struggled with tangent bundles and tangent vectors for that matter because I always think of them as associated with curves and not points.

Synthetic geometry stuff seemed to conceptually make more sense to me to a certain extent, but even there I struggled with the details.

Now that you seem to be trying to reconstruct standard concepts in the (more natural?) language of “higher stuff”, I wonder if I might have a chance to go back and reattempt to understand basic tangent bundles in this language.

First, would you say a regular tangent bundle is a tangle 0-bundle or a tangent 1-bundle? The degrees are a bit confusing. I would guess tangent 0-bundle because they are over points, but the relations to curves make me unsure (to say the least).

Another source of my confusion stems from my insistence to associate 1-forms with curves, i.e. in a way they are “dual” to curves. But they are also dual to tangent vectors. But tangent vectors are associated with points *gasp*.

I don’t know enough to enunciate my question clearly so I am relying on your ability to read my mind to make any sense out of this question :)

Eric

PS: I know you are otherwise extremely busy with more important things, so feel free to put this at the bottom of the “stack” :)

Posted by: Eric on June 14, 2007 4:52 PM | Permalink | Reply to this

### Re: Large Smooth Categories

I wonder if I might have a chance to go back and reattempt to understand basic tangent bundles in this language.

Sure!

would you say a regular tangent bundle is a tangent 0-bundle or a tangent 1-bundle?

Ordinary things are always 1-things. Or at least so it should be.

I would guess tangent 0-bundle because they are over points, but the relations to curves make me unsure (to say the least).

I see. Yes, so from this point of view it is good to look at the “integrated” version of these infinitesimal concepts.

For our purposes here, the best way to think of a tangent vector at a point is as a class of curves through that point, all sharing a certain property (that the derivative of any function along all these curves coincides at that point).

Curves are 1-dimensional beings. That’s not the reason why a tangent bundle is a tangent 1-bundle – but it is the reason why – as I claim – a good way to think of the tangent bundle is as a category (a 1-category, hence!): namely the path groupoid $P_1(X)$.

In a rather precise sense, the morphisms of $P_1(X)$ are essentially these curves that define tangent vectors.

associate 1-forms with curves, i.e. in a way they are “dual” to curves. But they are also dual to tangent vectors. But tangent vectors are associated with points

Well, a tangent vector is a point and the seed of a curve running through that point.

But what you address here is indeed an important issue. I like to put it in terms of this slogan:

infinitesimal = functorial + smooth

A smooth functor $P_1(X) \to \Sigma \mathbb{R}$ is a machine which reads in a curve in $X$ and spits out a number. But it does so smoothly and functorially.

Being smooth, the value of the functor may not wildly jump around as we vary the curve that we evaluate it on. It has to change in sufficiently “small steps”.

Being functorial means that the value of the functor on a curve from $x$ through $y$ to $z$ is already fixed once we know its value on the first part of the curve, from $x$ to $y$, as well as on the second part of the curve, from $y$ to $z$.

Now, putting these two properties, smoothness and functoriality, together, and imagining that we subdivide any given curve into more and more parts, it becomes heuristically clear that the value of the functor must in fact already be fixed by how it starts to change as we sit at a point $x$ and then start to run in some direction.

This heuristic idea may be turned into a theorem which says that smooth functors $t : P_1(X) \to \Sigma U(1)$ are “the same” as 1-forms $A$ on $X$. More precisely, for any such functor there is a 1-form $A$, such that the value of the functor on a curve $\gamma$ is nothing but the exponentiated integral of the 1-form $A$ along that curve

$t(x \stackrel{\gamma}{\to} y) = \exp \left( i \int_\gamma A \right) \,.$

That theorem is the key to understanding that the tangent bundle $T X$ of $X$ may, equivalently, be thought of as the category $P_1(X)$ of paths in $X$.

(Just for the record: John also talked about these things in his lecture, see for instance Quantization and Cohomology (Week 24)).

Posted by: urs on June 14, 2007 7:07 PM | Permalink | Reply to this

### Re: Large Smooth Categories

Last week I learned that ‘stack’ is “just” an external way to say: E-category. (A category with equality on morphisms, but not on objects.) E-categories are a standard way to encode categories in intensional type theory.

Durov talked about this at the MPI in Bonn. I guess it was this paper: http://arxiv.org/abs/0704.2030. Unfortunately, he spent almost all his time explaining standard Kripke–Joyal semantics. So this is all I know about his work.

What is interesting is that he is using Quillen model categories as internal objects. Whereas some research in type theory currently focuses on having it as models of intensional type theory.

So there seem to be two consecutive interpretations there.

Posted by: Bas Spitters on June 15, 2007 8:48 AM | Permalink | Reply to this

### Re: Large Smooth Categories

Bas Spitters wrote:

Last week I learned that ‘stack’ is “just” an external way to say: E-category. (A category with equality on morphisms, but not on objects.) E-categories are a standard way to encode categories in intensional type theory.

Interesting!

The logician Michael Makkai developed his theory of anafunctors in an attempt to ‘rid category theory of evil’. In category theory, it’s evil to assert equations between object. But, any functor involves lots of equations between objects:

$F(x) = y$

Makkai figured out a way around this. We don’t usually need a functor $F$ to give us a specific object $F(x)$ when we hand it an object $x$. We’re usually quite happy to get an object up to canonical isomorphism.

And, that’s often all we do get!

For example, consider the ‘tensor product functor’:

$\otimes : Vect \times Vect \to Vect$

We don’t really need this to give us a specific tensor product of vector spaces. We’re perfectly happy to get a vector space up to canonical isomorphism. Indeed, that’s all universal property of the tensor product gives us!

So, Makkai realized that the tensor product of vector spaces, and many other ‘functors’, don’t really become functors until after we make some arbitrary choices. Before that, they’re something else… they’re anafunctors!

You can always convert an anafunctor into a functor using the axiom of choice to pick a specific object $F(x)$ for each object $x$. But that’s just a sign that you’re doing something naughty.

Later, James Dolan and I developed a theory of $n$-categories based on Makkai’s idea. We used the term ‘virtual functor’ instead of ‘anafunctor’, but the concept was the same.

Later, Makkai developed a language for $\omega$-categories based on our work. In this language, you can’t say anything evil, since the concept of equality is completely gone. But, you can still say everything you really want to say.

Later still, when I was playing with smooth categories as an alternative to stacks, my student Toby Bartels realized that the 2-category

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

was the wrong one for this purpose. It’s much better to use

$[smooth categories, smooth anafunctors, smooth ananatural transformations]$

The reason, ultimately, is that the axiom of choice is simply false in the category of smooth spaces! When we accept anafunctors, all sorts of stacky ideas naturally follow.

So, it’s very interesting to see Nikolai Durov using ‘categories without equality between objects’ as a substitute for stacks!

Clearly lots of us coming from different directions are slowly creeping towards a big realization.

But the seeds of this realization are not new! They go back to the early days of topos theory, when people realized that topoi were good both for studying sheaves and for studying intuitionistic logic!

Great ideas come into the world as softly as doves. — Albert Camus

Posted by: John Baez on June 16, 2007 10:34 PM | Permalink | Reply to this

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