The n-Category Café

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?

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

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

than to think of

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?

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.

$\mathrm{Obj}(\mathrm{Vect})$ 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.

Let’s start by answering your questions […]

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

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

[…]

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

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

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

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

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

Personally, I don’t need a smooth structure on $\mathrm{Vect}$ 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.

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.

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

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.

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

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.

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

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.

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”).