## October 15, 2007

### What is the Fiber?

#### Posted by Urs Schreiber

I was involved in a discussion about how to best think of bundle gerbes, when introducing them to laypeople. Here the laypeople were supposed to understand what a fiber bundle is (unlike those complete laypeople to which we explained gerbes last time, when there were Gerbes in The Guardian).

The statement was made that like a sheaf is to a stack, so a principal line bundle is to a line bundle gerbe.

This worried me a little. I do think that, instead, bundle gerbes (as opposed to true gerbes!) should better be thought of as corresponding to transition functions:

$\array{ 1 & 2 \\ sheaf & stack \\ 1-cocycle & 2-cocycle \\ transition function & \mathbf{bundle gerbe} \\ principal bundle & principal 2-bundle }$

Maybe that’s an incredibly nitpicky, boring and irrelevant point. But I happen to think it is important. Here I will expand on it by:

- reviewing how we get the total space of a bundle from a transition function by first building a certain groupoid and then forming a certain pushout

- and how similarly we obtain the total 2-space of a 2-bundle from a bundle gerbe by first forming a certain two-groupoid and then doing a certain pushout.

I think of this as being an example of Toby Bartels’ general prescription for building a 2-bundle from transition data, as described in section 2.5.4 of his thesis 2-Bundles. But I will formulate it with a certain emphasis along the lines of my discussion with David Roberts at the end of this.

Ordinary bundles from bundles of groupoids obtained from their transition function

You all know how it works for ordinary bundles. But one of my points will be that instead of being thought of as a groupoid, the things we see when constructing bundle gerbes are actually best thought of as two groupoids. In order to make that point, I will first emphasize how the ordinary construction for bundles already uses 1-groupoids.

So suppose you have a principal $G$-bundle $P \to X$ which you have locally trivialized over a cover $\pi : Y \to X$ $t : \pi^* P \stackrel{\simeq}{\to} Y \times G$ to obtain a transition function $g : Y^{[2]} \to G \,.$

Over each point $y \in Y$ of the cover we have identitfied the fiber $P_x$ with $(y,G)$. Different points $y'$ covering the same point $x$ have the same typical fiber sitting over them, $(y',G)$, which will be identified with the former one by the action of the transition function $g(y,y')$.

This situation is modeled by a groupoid over each point $x$: it’s objects are the elements $h$ in the typical fibers corresponding to $y$

$(y,h)$

and there is a unique morphism from $(y,h)$ to that $(y',h')$ which is related by the action of $g(y,y')$:

$(y,h) \stackrel{g(y,y')}{\to} (y',h' = g(y,y') h) \,.$

Let’s call this groupoid $\tilde P_x := \left\{ (y,h) \stackrel{g(y,y')}{\to} (y',h' = g(y,y') h) | y \in \pi^{-1}(x)\,,h \in G \right\} \,.$

It’s a big fat groupoid which encodes all the information of our trivialization. But actually, this big fat groupoid is equivalent to a mere set:

to start with, it is non-canonically isomorphic to the set underlying $G$ itself $\tilde P_X \simeq G \,.$

This equivalence is established by picking any (and that’s the non-canonical choice) element $y_0$ in $\pi^{-1}(x)$ and using the cocycle to identify any morphism in $\tilde P_x$ with a morphism starting and ending at $y$. This equivalence thus involves naturality squares of the form

$\array{ (y,h) &\stackrel{g(y,y')}{\to} &(y',h' ) \\ \;\;\downarrow^{g(y,y_0)} && \;\;\downarrow^{g(y',y_0)} \\ (y_0,h_1) &\stackrel{\mathrm{Id}}{\to} &(y',h_2 ) } \,,$

where you can figure out $h_1$ and $h_2$ from the general rules. They are not important. What is important is that the squares of this kind always commute, precisely due to the cocycle property of the transition function $g$ (i.e. the fact that $g(y_1,y_2) g(y_2,y_3) = g(y_1,y_3)$).

This establishes the (non-canonical) equivalence of our groupoid with the typical fiber $G$ of the $G$-bundle that we started with.

But that typical fiber $G$ is itself non-canonically equivalent to the real fiber, $P_x$. In fact, if we remember the trivialization $t$ from above, then we find that our groupoid is indeed canonically equivalent to this fiber $\tilde P_x \simeq P_x \,.$

So, and that’s where I am getting to the point of the discussion with the bundle gerbes: we could say that the bundle of groupoids $\tilde P \to X$ which we have constructed from the transition function is the total space of the bundle which we started with.

But actually, that would be slightly imprecise. Rather, we would first want to apply this equivalence to reduce all the 1-groupoid fibers to mere sets.

My point is: a similar argument applies to bundle gerbes: while it is posible to regard the bundle gerbe itself as a 2-bundle, it is better thought of as just the puffed-up thing analogous to $\tilde P$ which we obtain from the transition function of a 2-bundle, and to address as a true 2-bundle only the result of operating with some quotienting operation on our bundle gerbe.

Before closing this first part, I’ll reformulate the above in the language of universal bundles using groupoids, as mentioned at the end of this

Writing $Y^{[2]}$ for the groupoid associated to $Y \to X$, writing $\Sigma G$ for the one-object groupoid obtained from the group $G$ (it is crucial here to distinguish $G$ from $\Sigma G$, otherwise we’ll get lost) and $G//G = \mathrm{INN}(G)$ for the action groupoid of $G$ acting on itself by left action, we have:

the transition function itself is a functor

$Y^{[2]} \stackrel{g}{\to} \Sigma G \,.$

Along this functor we pull back the universal $G$-bundle in its groupoid incarnation $\array{ && \mathrm{INN}(G) \\ && \downarrow \\ Y^{[2]} &\stackrel{g}{\to}& \Sigma G }$ to obtain the bundle of groupoids $\tilde P$ $\array{ \tilde P &\to& G // G \\ \downarrow && \downarrow \\ Y^{[2]} &\stackrel{g}{\to}& \Sigma G } \,.$ The orginal $G$-bundle we started with is reobtained as the pushout $\array{ \tilde P &\stackrel{t}{\to}& Y \times G \\ \downarrow^s && \downarrow \\ Y \times G &\to& P } \,.$

2-bundles from bundles of 2-groupoids obtained from their transition 2-function

For our line 2-bundle the transition function is now a pseudofunctor

$Y^{[2]} \to \Sigma (\Sigma U(1))$ mapping a triangle $\array{ && y' \\ & \nearrow && \searrow \\ y &&\to&& y'' }$

to a filled triangle

$\array{ && \bullet \\ & \nearrow &\;\,\;\downarrow^{g(y,y',y'')}& \searrow \\ \bullet &&\to&& \bullet } \,.$

Again we form the pullback along $\array{ && \mathrm{INN}_0(\Sigma U(1)) \\ && \downarrow \\ Y^{[2]} &\stackrel{g}{\to}& \Sigma \Sigma U(1) }$

to obtain $\tilde P$. This is now a bundle of 2-groupoids!

A typical 2-morphism here looks like

$\array{ && y' \\ & \multiscripts{^h}{\nearrow}{}\; &\;\,\;\Downarrow^{g(y,y',y'')}& \;\searrow^{h'} \\ y &&\stackrel{h''}{\to}&& y'' } \,,$ where $h'' = g(y,y',y'')h h' \,,$ for all $y$, $y'$ and $y''$ and all all $h$ and $h'$.

To see the bundle gerbe here consider this as the “Hitchin-Chatterjee” version of bundle gerbes, where we have assumed the transition line bundle $\array{ L \\ \downarrow \\ Y^{[2]} }$ to be trivial. Alternatively, replace the group elements on the edges by elements of $U(1)$-torsors or by elements of complex 1-dimensional vector spaces, if you like. Then $g(\cdots)$ is the component map of the bundle gerbe multiplication mormphism $\mu_g : \pi_1^*(L) \otimes \pi_2^*(L) \to \pi_3^* L \,.$ Notice that there are three different ways to interpret this associative product:

a) as defining a central extension of the 1-groupoid $Y^{[2]}$ (that’s the point of view found in much of the literature)

b) as a $U(1)\mathrm{Tor}$ or $1d\mathrm{Vect}$-enriched groupoid (that’s pretty obvious and will probably strike few people as being really different from a))

c) as a two-groupoid where horizontal composition is just the tensor product, and where the “multiplication” is only given by the 2-morphisms.

Recall how we already had a 1-groupoid $\tilde P$ for the 1-bundle case. From this point of view we clearly want to be thinking of $\tilde P$ now here as a 2-groupoid.

But the point is: like our 1-groupoid before was equivalent to the typical fiber (a set, aka 0-groupoid) our 2-groupoid here is equivalent to a 1-groupoid, namely to $\Sigma U(1) \,.$

And that’s good! Because that is indeed the typical fiber we expect for a line 2-bundle.

To see the equivalence, use the cocycle condition, in complete analogy to what we did before:

we want to build a commuting triangular cylinder whose top triangle is

$\array{ && y' \\ & \multiscripts{^h}{\nearrow}{}\; &\;\,\;\Downarrow^{g(y,y',y'')}& \;\searrow^{h'} \\ y &&\stackrel{h''}{\to}&& y'' }$

and whose bottom triangle is of the form

$\array{ && y_0 \\ & \multiscripts{^{h_1}}{\nearrow}{}\; &\;\,\;\Downarrow^\mathrm{Id}& \;\searrow^{h_2} \\ y_0 &&\stackrel{h_3}{\to}&& y_0 } \,.$

That’s a simple exercise. I am too tired to try to fake drawing the diagrams here. The point is: it can be done naturally and smoothly, simply by filling everything with transitions in the only obvious way.

So, this way we find that our bundle gerbe, which we should really think of as giving rise, directly, to a bundle $\tilde P$ of 2-groupoids, becomes equivalent to the bundle of 1-groupoids which we expect, namely a bundle of groupoids each of which are equivalent to $\Sigma U(1)$.

Bottom line slogan

A bundle gerbe “is” a 2-transition function. From any $n$-transition function we obtain, by a pullback operation, a bundle of $(n+1)$-groupoids. The bundle of $n$-groupoids which we identify as the total space of the $n$-bundle associated with our transition function is obtained by performing a certain pushout on that.

Hm.

Rereading this, I am worried that everybody will think how completely pointless this is. But I wanted to write it down somewhere. I think it is both important for putting bundle gerbes into perspective and, more importantly, it helps understand the concept of principal 2-bundles and their construction from transition functions, the way Toby Bartels describes it in his thesis (check the proof of theorem 22!) or the way I think it can equivalently be reformulated in terms of pullback of the universal tangent groupoid sequence $\mathrm{INN}_0(G_{(2)}) \to \Sigma G_{(n)}$.

By the way, Toby tells me that the most up-to-date version of his thesis is provided here.

Posted at October 15, 2007 9:11 PM UTC

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### Re: What is the Fiber?

I’ve often thought that a double groupoid might be an appropriate device to whip out at this moment, or for $n$-bundles, a groupoid internal to $n$-categories.

Back to 2-groupoids, though: I’m not sharp enough at the moment, but I think you’re saying the 2-groupoid has as space of objects the base space you’re interested in? We need to sort this one out, because claiming $INN_0(G_2)$ is the universal 2-bundle needs a little more evidence, in my opinion (saying this because I know my supervisor will ask …difficult questions ;-)

[Also, some of the MathML didn’t come out right in your post, Urs - missing a slash on a “downarrow” command.]

Posted by: David Roberts on October 16, 2007 6:39 AM | Permalink | Reply to this

### Re: What is the Fiber?

I’ve often thought that a double groupoid might be an appropriate device

Lately, the gods keep sending me messangers that point me to $n$-fold groupoids.

All I need to do is figure out if they are just trying to tease me. :-)

I think you’re saying the 2-groupoid has as space of objects the base space you’re interested in?

In the case where the structure (2-)group is $\Sigma U(1)$, which has just a single object, the objects of the pullback 2-groupoid $\tilde P$ $\array{ \tilde P &\to& \mathrm{INN}_0(\Sigma U(1)) \\ \downarrow && \downarrow \\ Y^{[2]} &\to& \Sigma(\Sigma U(1)) }$

are just those of $Y$. Yes.

But more generally, when $G_{(2)}$ is the strict 2-group coming from the crossed module $H \stackrel{t}{\to} G \stackrel{\alpha}{\to} \mathrm{Aut}(H)$, then, since $\mathrm{INN}_0(G_{(2)})$ has space of objects $G$, it follows that the pullback

$\array{ \tilde P &\to& \mathrm{INN}_0(\Sigma G_{(2)}) \\ \downarrow && \downarrow \\ Y^{[2]} &\to& \Sigma(G_{(2)}) }$

has objects which are pairs $(y,g) \,,$

with $y$ an element of $Y$ and $g$ any object of $G$.

For fixed $y$, these objects are to be thought of as being the group of objects of the typical fiber (which is $G_{(2)} = (H \to G)$) sitting over $y$.

a little more evidence

From one point of view all that seems to remain is the proof of local trivializability of the realization. You said you thought about it. Did you solve it?

But from another point of view, using transport and don’t passing to realizations, my impression is that it boils down to a triviality.

But also trivialities need to be spelled out. When I find the time I get back to you about this.

Posted by: Urs Schreiber on October 16, 2007 11:44 AM | Permalink | Reply to this

### Re: What is the Fiber?

Also, some of the MathML didn’t come out right

Thanks! Fixed.

Posted by: Urs Schreiber on October 16, 2007 11:54 AM | Permalink | Reply to this

### Translation into the language of Brylinski

Urs, I think you and I come from slightly different angles on this gerbes thing. It would be most helpful for me if you could translate your thinking into the language of Brylinski. Then I could compare “ontological concepts” on the same level, and I’d be in a better position to understand the bigger picture.

So here’s the challenge, the elephant we are going to try and sketch : the gerbe defined in Brylinski page 194. I would like you to express for me all the various ways you have of thinking about this geometric beast. I will recall the example here, and then we can compare notes.

For the unitiated, Brylinski thinks of a gerbe on a space $X$ as a sheaf of groupoids on $X$. To be precise, let’s talk about “unitary Dixmier-Douady gerbes” - these are gismos such that the hom-sets of the groupoid they assign to some small neighbourhood $U$ are $\overline{U}(1)$-torsors, where $\overline{U}(1)$ is the sheaf of functions on $U$ with values in $U(1)$ (underlining is not allowed at the n-cafe ;-)). And when I say “sheaf on $X$”, I really mean a stack over the site of smooth manifolds mapping into $X$.

Let $1 \rightarrow U(1) \rightarrow \tilde{G} \rightarrow G \rightarrow 1$ be a central extension of Lie groups, and $P \rightarrow X$ a principal $G$-bundle. From this data, we’re going to construct a gerbe on $X$, which measures the obstruction to finding a principal $\tilde{G}$-bundle $Q \rightarrow X$ such that $Q/U(1) \cong P$.

In other words, we’ve been told how to extend $G$ at a point ; and we’re asking if we can smoothly vary this extension over $X$ consistently.

The gerbe is defined as follows. Recall I like to do things from the point upwards - what stacks/sheaves assign to $\pt \rightarrow X \in Man$ are the fundamental entities, the rest is just telling us what it means to be “smooth”.

What this gerbe assigns to an object $pt \rightarrow X$ in our site - i.e. to a point $x \in X$ - is simply the groupoid whose objects are pairs $(T, \phi)$ where $T$ is a $\tilde{G}$-torsor and $\phi : T/U(1) \stackrel{\cong}{\rightarrow} P_x$ is an isomorphism of $G$-torsors. The morphisms in the groupoid are obvious.

What the gerbe assigns to objects $U \rightarrow X$ in our site is then just the “$U$-parametrized” version of the above.

Okay, that’s the gerbe. Now for the things which confuse me:

1. How is this gerbe thought of in the bundle-gerbe picture, the principal 2-bundle picture, the 2-cocycle picture, and the other pictures?

2. What is the fundamental geometric object? Is Brylinski’s gerbe just the “shadow” of some more fundamental geometric description?

3. Here’s another big confusion of mine : vaguely, gerbes are supposed to be the geometric objects representing classes in $H^2(X, U(1))$, the analogue of the way line-bundles represent classes in $H^1(X, U(1))$. But… there are two acceptable ways to “categorify” line bundles, and they lead to different geometric descriptions!

On the one hand, you might argue that a line bundle is “something which assigns a 1d vector space to each point of $X$”. Thus a gerbe should be “something that assigns a 1d 2-vector space to each point of $X$”.

On the other hand, you might argue that a line bundle is “something which assigns a set to each point of $X$”. Thus a gerbe should be “something that assigns a groupoid to each point of $X$”. (Here I’m using the reasonable idea that groupoids categorify sets).

How are these pictures related? Is the 2-line-bundle the representation category of the groupoid-bundle?

Posted by: Bruce Bartlett on October 17, 2007 1:28 AM | Permalink | Reply to this

### Re: Translation into the language of Brylinski

Here is the quick answer, more details when I have more time:

Given a bundle $P$ its sheaf of sections is the sheaf of local trivializations

$U \mapsto \mathrm{Hom}(U \times G, P|_U) \,.$

Similarly, given a 2-bundle $P$ its stack of sections should be the stack of local trivializations

$U \mapsto \mathrm{Hom}(U \times G_{(2)}, P|_U)$

Since any transformation between two homomorphisms of $G_{(2)}$-torsors is invertible, these Hom-spaces are groupoids. So this is a stack in groupoids, called a gerbe.

You can find this spelled out explcitly for bundle gerbes in a paper by Michael Murray which I will dig out for you as soon as I find the time. There is is shown that the local trivializations of a bundle gerbe constitute the corresponding stack of groupoids.

Your last question is simpoly about principal versus associated picture: the ultimate distinction is:

a bundle has fibers that are objects in a 1-category. A 2-bundle has fibers that are objects in a 2-category.

Posted by: Urs Schreiber on October 17, 2007 11:04 AM | Permalink | Reply to this

### Re: Translation into the language of Brylinski

Similarly, given a 2-bundle P its stack of sections should be the stack of local trivializations…this is a stack in groupoids, called a gerbe.

This is the crucial point : in your picture, the sheaf of groupoids is the space of sections of a more fundamental geometric object - the principal 2-bundle. But are you sure that is the correct viewpoint? The alternative is that the sheaf of groupoids is the fundamental geometric object, and the principal 2-bundle is… well I don’t know what it is :-)

What is the principal 2-bundle description of the gerbe above (the one coming from central extensions of Lie groups)? That is what I’d really like to know. Perhaps this is well-known… but I had better understand it now, so that I can get my “viewpoint” correcly aligned.

Posted by: Bruce Bartlett on October 17, 2007 11:37 AM | Permalink | Reply to this

### Re: Translation into the language of Brylinski

Sorry to make yet another post :-) But I am eager to get to the bottom of this, because it has been confusing me for ages.

I think we still have very different ideas of what stacks and sheaves are .

Let’s consider a line-bundle $L$ on a space $X$. I think we’d both agree that, however one phrases it, the “fundamental geometric object” is that thing which assigns to each point in $x$ the fiber sitting above it:

(1)$x \mapsto L_x.$

Another way to view $L$ is via its sheaf of sections over open sets of $X$

(2)$\mathcal{L}^' : Open Sets(X) \rightarrow \mathbb{R}_X-modules.$

I’ve given it a prime because somehow this is the old-fashioned/pre-site idea of sheaves as living over the open sets of $X$. I think we both agree that this viewpoint is sort of secondary.

But there’s another “stacky” way to think of $L$ - as a sheaf $\mathcal{L}$ over the site $Man/X$. This is my favourite view, and I would argue that it is fundamental.

This is the sheaf which assigns to an object $pt \rightarrow X$ in $Man/X$ - in other words, a point $x \in X$ - the “space of sections of the pullback bundle”… in other words, the fiber!

(3)$x \mapsto L_x$

We’ll call this sheaf $\mathcal{L}$, which lives over $Man/X$, the “stacky” sheaf. The beauty about stacky sheaves is that they really do fit our geometric intuition : they assign things to points in $X$.

Now let’s go up one dimension. A Brylinski-gerbe is a stack over $Man/X$… in other words, it is a smooth assignment of a groupoid $\mathcal{E}_x$ to every point $x \in X$. This is all we care about - the rest simply defines “smooth assignment”, and we only need it when the auditors come around.

So in this picture, there is every reason to suspect that the Brylinski gerbe is the “principal 2-bundle-ish thing”… it is something which assigns geometric things (groupoids) to points in $X$. It doesn’t seem to be the space of sections of some more fundamental entity.

Posted by: Bruce Bartlett on October 17, 2007 12:15 PM | Permalink | Reply to this

### Re: Translation into the language of Brylinski

the Brylinski gerbe is the “principal 2-bundle-ish thing”

I agree completely, that this is a very important point (for those, like us, who care about such hair-splitting): when we see groupoids being assigned to soemthing, we need to try to be careful whether what we are seeing are groupoids of sections or rather fibers. And care is needed since over a point these two concepts are indistinguishable.

I must admit that I haven’t looked closely enough at Brylinksi to decide this. Possibly his construciton is really better thought of as defining the smooth groupoid that the bundle gerbe people also consider, where smoothness is encoded by realizing the groupoid internal to sheaves, hence by having a sheaf of groupoids.

You have Brylinski in front of you and you have my consent that you are looking at things from the right perspective, so you can figure it out. I need to think about soemthing else right now. :-)

Posted by: Urs Schreiber on October 17, 2007 3:45 PM | Permalink | Reply to this

### Re: Translation into the language of Brylinski

But are you sure that is the correct viewpoint?

I think that the right point of view, as far as that makes sense, is the one that generalizes straightforwardly. Of course that depends to some degree on your applications.

I would say: compare the number of lines you need to define a connection on a nonabelian $n$-thing to see which point of view does what.

The alternative is that the sheaf of groupoids is the fundamental geometric object, and the principal 2-bundle is

Assuming that we agree that it is hard to argue that this question is about anything else but about subjective taste, let me say:

if you ask me, I am convinced that the fundamental picture is neither $n$-sheaves of sections nor total $n$-spaces of $n$-bundles, but – $n$-transport.

Even in the absence of connections, i.e. when we work just on the “category” of constant paths, I think it is good to think of an $n$-bundle as a (smoothly locally trivializable) fiber-assigning functor $X \to T$ for $T$ the desired $n$-category of fibers.

What is the principal 2-bundle description of the gerbe above (the one coming from central extensions of Lie groups)?

Different ways to do that. If you are happy with thinking of a bundle gerbe as a 2-bundle (along the lines of the above entry), then the standard way is this:

take your $G$-bundle $P$, form $P \times_X P$ from which you have a canonical map $P \times_X P \to G$ and pull back along this map the $U(1)$-bundle which is the central extension $\hat G \to G$ to obtain $\array{ L &\to& \hat G \\ \downarrow && \downarrow \\ P^{[2]} &\stackrel{\pi}{\to}& G } \,.$

That defines a bundle gerbe for you. So the fibers over $X$ are the groupoids $\pi^{-1}(x)$ centrally extended by $L$, which are equivalent to the structure 2-group $\Sigma U(1)$.

People who like to think of gerbes entirely in terms of $PU(H)$-bundles will in fact think of the original $PU(H)$-bundle itself already as being the gerbe. That’s morally justified by the reasoning mentioned in that other comment: $PU(H)$ is the automorphism group of the algebra $K(H)$, hence you can canonically associate a $K(H)$-bundle and think of the algebra fibers as really being the corresponding module categories (= 1-dimensional 2-vector spaces).

It is a pity that Konrad and my “Parallel 2-Transport and 2-Functors” isn’t there yet.

Posted by: Urs Schreiber on October 17, 2007 3:38 PM | Permalink | Reply to this

### Re: Translation into the language of Brylinski

Is the 2-line-bundle the representation category of the groupoid-bundle?

Let me elaborate on this. I’ve always had a fundamental “obstruction” (excuse the pun) to understanding higher geometry : there is a school of thought which says that a gerbe is a 2-line bundle on a space $X$. But I’ve never seen a geometric example!

By ‘geometric example’, I mean one where you are associating to each point $x \in X$ a 1d 2-vector space in a natural way. ‘Calculational models’, like higher gauge theory don’t (yet) do this - there one simply assumes the line-bundle is trivial… and you ignore it completely, and concentrate on its structure 2-group. Something like that.

But observe that if you do follow the approach I threw out above, and pass from Brylinski-style gerbes to 2-line-bundle-style gerbes (by taking the representation categories of the fibers), then you do indeed get natural “geometric” 2-line bundles!

For instance, here’s the 2-line-bundle in the previous example: it assigns, to each point $x \in X$, the category of representations of the groupoid $\mathcal{E}_x$ (the groupoid we are assigning to each point $x \in X$).

Recall that $\mathcal{E}_x$ was defined to be the groupoid whose objects were pairs $(T, \phi)$ where $T$ is a $\tilde{G}$-torsor and $\phi : T/U(1) \rightarrow P_x$ is a torsor isomorphism. A morphism $\theta : (T, \phi) \rightarrow (T^', \phi^')$ is an isomorphism of $\tilde{G}$-torsors $\gamma : T \rightarrow T^'$ such that the obvious diagram commutes.

So the 2-line bundle is going to be assigning

(1)$x \mapsto Rep(\mathcal{E}_x)$

(By “Rep” I obviously mean unitary representations… are there any other kind?). Now the groupoid $\mathcal{E}_x$ is connected, and its hom-sets are $U(1)$-torsors, so it’s equivalent to the group $U(1)$. Thus the fibers of our 2-line bundle will look like the representation category of $U(1)$.

But the fibers are naturally associated to each point $x \in X$…which I find an interesting construction. Is it good for anything?

Posted by: Bruce Bartlett on October 17, 2007 11:24 AM | Permalink | Reply to this

### Re: Translation into the language of Brylinski

But I’ve never seen a geometric example!

You have:

the 1-dimensional 2-vector space is $\simeq \mathrm{Mod}_{\mathbb{C}} = \mathrm{Vect}$, where the equivalence is in $\mathrm{Bim} \hookrightarrow 2\mathrm{Vect}$. Since compact operators are Morita equivalent to the ground field, $K(H) \simeq \mathbb{C}$, the 2-bundle whose fibers are the 1-dimensional 2-vector spaces $\mathrm{Mod}_{K(H)}$ is nothing but an ordinary bundle of compact operators. That you obtain canonically associated to a $PU(H)$-bundle. And those are indeed the same as bundle gerbes.

String 2-bundles are another example of this kind, built from bundles of vN algebras instead.

Recall the table in 2-Groups and Algebras and the discussion in 2-Vector transport and Libe Bundle Gerbes.

More generally, take Toby’s theorem 22 to see a total 2-space of any associated 2-bundle given its cocycle data.

Posted by: Urs Schreiber on October 17, 2007 3:22 PM | Permalink | Reply to this

### Re: Translation into the language of Brylinski

Brylinski’s “Categories of vector bundles and Yang-Mills equations” in Contemporary Mathematics no. 230 will, I feel, address some of the points you raise. He talks about 2-vector bundles, but really only for rank one, and relates them to his previous work. He uses KV 2-vector spaces, but takes a little while to get there, doing a whole lot of geometry first.

I was unaware of this article for a long time, until Michael Murray gave me a photocopy of his photocopy - our library being too stingy to keep up our subscription.

Posted by: David Roberts on October 18, 2007 3:26 AM | Permalink | Reply to this

### Re: Translation into the language of Brylinski

Thanks, David. I hope our library’s got it! I’ve never read anything of Brylinski beside the book, though I did watch a video of a talk he gave at the MSRI. I’m a big fan though.

Yesterday was a red letter day in my understanding of gerbes. Amongst other things, I finally got up to speed on bundle gerbes. I’ve thought over what Urs has said, and I have many comments to make, the majority of which I’ll make via email.

But the following point is very important. I guess I’m addressing Urs now :-)

If you ask me, I am convinced that the fundamental picture is neither $n$-sheaves of sections nor total $n$-spaces of $n$-bundles, but $n$-transport.

Even in the absence of connections, i.e. when we work just on the “category” of constant paths, I think it is good to think of an n-bundle as a (smoothly locally trivializable) fiber-assigning functor

This is precisely the point of confusion. I agree with you completely - the fundamental geometric object is the thing which assigns fibers to points $x \in X$.

But that’s what a stack is!

It’s a false trichotomy. There are only two pictures in the game : top-down (total spaces) and bottom-up (fiber assigning things).

You think of a stack/sheaf as some kind of “sections device”. Indeed it is, but “sections device” = “fiber assigning functor”. (I know you know this, and I apologize for harping on about it. Consider the following as a rant to myself.)

To see this, you have to work in the site language. That’s the great unification step. The amazing thing about a sheaf over $Man/X$ - as opposed to an old-fashioned sheaf over Open_sets$(X)$ - is that you can talk about points .

Remember, I’m thinking of sheaves/stacks bottom-up, i.e. functors(2-functors) from the site into Set(Gpds).

A sheaf $\mathcal{F}$ over $Man/X$ is, at its very core, something which smoothly assigns a fiber to every point $x \in X$. This is the value of the sheaf on the objects $pt \stackrel{x}{\rightarrow} X$ in our site:

(1)$x \mapsto \mathcal{F} (pt \stackrel{x}{\rightarrow} X).$

That’s the fundamental information of a sheaf/stack. The rest (what it does to other objects in the site) is just there to define the meaning of “smoothly assigns”. It’s secondary.

In other words, taking sections at a point is the same as assigning a fiber to that point. I’m in the same boat as you - we both like to think in terms of n-transport!

I just realized that, perhaps the 2-bundle picture is precisely the same as the gerbes picture.

I guess in the principal 2-bundle picture, we want to assign to each point of $X$ a 2-torsor for the the 2-group $\Sigma^2 U(1)$.

But isn’t it true that a 2-torsor for $\Sigma^2 U(1)$ is the same thing as a connected groupoid whose hom-sets are $U(1)$-torsors? That would be cool!

Because then, the 2-bundle picture and the Brylinski picture are tautologically the same. In the Brylinski picture, a gerbe is simply something which assigns a groupoid (connected and whose hom-sets are $U(1)$-torsors) to every point of $X$:

(2)$x \mapsto \mathcal{E}_x$

Isn’t this the same thing as saying they are 2-torsors for the 2-group $\Sigma^2 U(1)$? To prove this, remember Michael Murray’s point (see I did learn about bundle gerbes :-)) about categories whose hom-sets are $U(1)$-torsors : they must be groupoids, because the $U(1)$-torsor condition means $hom(a,b)$ must match up with $hom(b,a)$.

Posted by: Bruce Bartlett on October 18, 2007 11:17 AM | Permalink | Reply to this

### Re: Translation into the language of Brylinski

I just realized that, perhaps the 2-bundle picture is precisely the same as the gerbes picture.

Indeed, this is Toby’s Last Theorem, though he conceives of it slightly differently.

Let me sum-up my position : everyone is agreed (I even include bundle-gerbey people in this) that a $U(1)$-gerbe on a space $X$ is a smooth assignment of a groupoid $\mathcal{G}_x$ to every point of $X$:

(1)$x \mapsto \mathcal{G}_x$

These groupoids must have the property that they are connected and their hom-sets must be $U(1)$-torsors. Said differently, they are 2-torsors for the 2-group $\Sigma^2 U(1)$.

The only difference is how we implement the notion of “smooth assignment”.

Gerbey people like me interpret “smooth assignment” in the language of stacks. Principal-2-bundley people do it in the language of open covers. These are ultimately the same, it’s a matter of taste.

Posted by: Bruce Bartlett on October 18, 2007 12:24 PM | Permalink | Reply to this

### Re: Translation into the language of Brylinski

Gerbey people like me interpret “smooth assignment” in the language of stacks. Principal-2-bundley people do it in the language of open covers.

I forgot to add : Bundle-gerbey people interpret “smooth assignment” by ensuring that the groupoids $\mathcal{G}_x$ which appear are always living in some big manifold $P$, where it becomes obvious what “smooth” means.

Posted by: Bruce Bartlett on October 18, 2007 12:31 PM | Permalink | Reply to this

### Re: Translation into the language of Brylinski

I just realized that, perhaps the 2-bundle picture is precisely the same as the gerbes picture.

Hold on. Gerbes or bundle gerbes?

Gerbes is “section assignment”

$U \mapsto \Gamma(P|_U) \,.$

2-bundles means “total space”

$P \to X$

or fiber-assignment

$x \mapsto P_x \,.$

Bundle gerbes is (essentially at least) the total space picture.

But I am not sure yet that we really want to go as far as saying “section assignment”

$U \mapsto \Gamma(P|U)$

is “precisely the same” as fiber assignment

$x \mapsto P_x \,.$

Posted by: Urs Schreiber on October 18, 2007 1:23 PM | Permalink | Reply to this

### Re: Translation into the language of Brylinski

Hold on. Gerbes or bundle gerbes?

Gerbes.

Gerbes is “section assignment” $U \mapsto \Gamma(P|_U)$.

No, that is just “part (c)” (i.e. the not-so-important-part) of what section assignment means. Firstly, one should work over the site $Man/X$ and not Open_sets$(X)$. Then a gerbe is section assignment of the pullback bundle:

(1)$(U \stackrel{f}{\rightarrow} X) \mapsto \Gamma(f^* P).$

The most important information - the “part (b)” information - is what it does to points $(pt \stackrel{x}{\rightarrow} X)$, where it just assigns the fiber:

(2)$x \mapsto P_x.$

As for bundle gerbes, they are “ontologically” on the same level as gerbes (I disagree with you that they are categorifications of transition functions). In other words, they are also gadgets which smoothly assign groupoids $P_x$ to points in $x$:

(3)$x \mapsto P_x.$

To interpret “smoothly assign”, they simply restrict themselves to those situations where each groupoid $P_x$ actually lives in some huge manifold, in which case “smoothly assign” means the obvious thing.

Posted by: Bruce Bartlett on October 18, 2007 2:29 PM | Permalink | Reply to this

### Re: Translation into the language of Brylinski

As for bundle gerbes, they are “ontologically” on the same level as gerbes (I disagree with you that they are categorifications of transition functions).

Let $Y \to X$ be a surjective submersion.

This is a transition function with values in $G$ wrt $Y$:

$Y^{[2]} \to \Sigma G \,,$

this is a principal bundle gerbe wrt $Y$:

$Y^{[2]} \to \Sigma U(1)\mathrm{Tor} \,,$

this is a line bundle gerbe wrt $Y$:

$Y^{[2]} \to \Sigma 1d\mathrm{Vect} \,.$

From the transition function we can easily rebuild a bundle of sets. From the bundle gerbe we can rebuild a bundle of groupoids.

The rebuilding process is an important step. I think you are choosing to regard it as given for granted. And in fact much of the literature on bundle gerbes does so. I would like to keep it explicit.

Apart from that I very much agree with the rest you say, up to the following extra remark:

when you identify “smooth fiber assignment” with “section assignment” you are thinking of the case where the statement

sections over a point are equivalent to the fiber $\Gamma(P|_x) \simeq P_x$

holds.

This is often true. Sections are morphisms into fibers. Over a point they are the same as fibers if the fibers coincide with their generalized elements.

But it is useful and important to have the notion of section more generally.

A line bundle gerbe is a functor with values in 1d vector spaces, $Y^{[2]} \to \Sigma 1d\mathrm{Vect}$. It may not have a global section wrt to $1d\mathrm{Vect}$. But if it is torsion it has a global section wrt the larger context $\Sigma \mathrm{Vect} \supset \Sigma 1d\mathrm{Vect} \,,$ meaning that there are morphisms into it when we embed in the larger context

$\array{ &&\Sigma \mathrm{Vect} \\ {}^{\mathrm{Id}_{\mathbb{C}}}\nearrow&\Downarrow^e & \uparrow \\ Y^{[2]} &\to& \Sigma 1d\mathrm{Vect} } \,.$

This $e$ is a twisted vector bundle, as you know.

But similar considerations already apply for ordinary bundles: a vector bundle may have ordinary sections, non-vanishing sections, holomorphic sections, sections of finite support, etc.

I am just making a trivial point here: we want to keep “section assignment” and “fiber assignment” as two different concepts, even though they may actually come pretty close to coinciding in many cases.

This point is independent of all smoothness considerations. It applies to the world of $n$-bundles over finite sets, too.

Posted by: Urs Schreiber on October 18, 2007 2:59 PM | Permalink | Reply to this

### Re: Translation into the language of Brylinski

I have many comments to make, the majority of which I’ll make via email

The amazing thing about a sheaf over $\mathrm{Man}/X$ - as opposed to an old-fashioned sheaf over $OpenSets(X)$ - - is that you can talk about points .

Okay. Good.

So let’s see if we agree:

there are, after all, three points of view:

a) total space with projection $P \to X$

b) fiber assigning map $x \mapsto P_x$

c) section-assigning map $U \mapsto \Gamma(P|_U)$

You emphasize (as in our previous elephant-and-pizza discussion) that one way to say how b) is smooth is to conceive the fiber-assigning-functor as a stack. That’s great. I appreciate this point of view.

But now lets see: does that mean that b) and c) coincide? Just because both involve sites/stacks etc. doesn’t mean that they are “the same”. I’d still think both exist independently.

In fact, I’d think that after you have went ahead and conceived b) in terms of a sheaf on $\mathrm{Man}/X$, you can still, on top of this picture then, consider c) in that context. It would involve something like for each $U$ a sheaf on $\mathrm{Man}/U$ and a map of sheaves from $\mathrm{Man}/U$ to $\mathrm{Man}/X$ etc. and so on. In other words, you’d interpret the schematic formula $U \mapsto \Gamma(P|_U)$ entirely in the world of families over manifolds in order to encode this way what smoothness means here.

So I really think points a), b) and c) are separate concepts, and that there are (at least possibly) two ways to say what it means for a), b) and c) to be smooth:

- either we work with categories internal to sheaves over manifolds (what John and I have mostly been doing)

- or we work with stacks over manifolds (the point of view you like so much).

But isn’t it true that a 2-torsor for $\Sigma^2 U(1)$ is the same thing as a connected groupoid whose hom-sets are $U(1)$-torsors? That would be cool!

Great. Now we are exactly at the point that my entry above was supposed to address.

(By the way, you mean $\Sigma U(1)$ instead of $\Sigma^2 U(1)$ The latter is a 3-group.)

I am saying: the answer to your question is YES and this is indeed cool, IF we pay close attention to some subtlety involving the categorical dimension:

Let’s face it: a “groupoid whose hom-spaces are $U(1)$-torsors” is really:

- either a groupoid enriched over $U(1)\mathrm{Tor}$

- or, equivalently, the graph (in the sense of graph of functions) of a pseudofunctor from a codiscrete groupoid to $\Sigma U(1)\mathrm{Tor}$.

If we choose the first point of view, we need to switch entirely into the enriched world for the rest of our discussion. It is more straightforward for our purposes to adopt the second point of view here.

That means, we regard a “groupoid whose hom-spaces are $U(1)$-torsors” really as a 2-groupoid

- with the same objects as expected

- whose morphsims

$a \stackrel{T_{a b}}{\to} b$ consist of a pair of objects, $a$, and $b$ and a morphism $T_{a b}$ in $\Sigma U(1)\mathrm{Tor}$

composition of such 1-morphisms is the tensor product of $U(1)$-torsors

$a \stackrel{T_{a b}}{\to} b \stackrel{T_{b c}}{\to} c \;\; = \;\; a \stackrel{T_{a b} \otimes_{U(1)} T_{b c}}{\to} c$

- whose 2-morphisms

$\array{ && b \\ & {}^{T_{a b}}\nearrow & \Downarrow^{g_{a,b,c}}& \searrow^{T_{b c}} \\ a &&\stackrel{T_{a c}}{\to}&& c }$

are 2-morphsism $T_{a b} \otimes_{U(1)} T_{b c} \stackrel{g(a,b,c)}{\to} T_{a c}$

in $\Sigma U(1)\mathrm{Tor}$, namely $U(1)$-torsor homomorphisms.

Their composition is the composition inherited from $\Sigma U(1)\mathrm{Tor}$ and they are required to make the obvious tettrahedra with respect to the underlying codiscrete groupoid on the objects 2-commute.

Such 2-groupoids are indeed torsors over $\Sigma U(1)$: the single object of $\Sigma U(1)$ acts trivially on them, and the morphism $c$ of $\Sigma U(1)$ yields the transformation between two trivial actions (between the identity 2-functor and itself) whose components map is

$a \stackrel{T_{a b}}{\to} b \;\; \mapsto \;\; \array{ a &\stackrel{T_{a b}}{\to}& b \\ {}^{U(1)}\downarrow &\Downarrow^{\cdot c}& \downarrow^{U(1)} \\ a &\stackrel{T_{a b}}{\to}& b }$

Notice that this a weak transformation: its component map is just a pseudofunctor.

(We can get rid of all the pseudo here, if we like, by throwing in more generators and relations. This is the standard issue that comes up in this context everywhere. It is related to how you think of a 2-cocycle with values in a 2-group $G_{2)}$: either it is a pseudofunctor $Y^{[2]} \to \Sigma G_{(2)}$ or a proper 2-functor, whose domain is a refinedment of $Y^{[2]}$ to a 2-category).

Because then, the 2-bundle picture and the Brylinski picture are tautologically the same.

Well, that’s what my entry was about: to which extent provides the bundle gerbe picture already a 2-bundle picture. Brylinski’s picture is the bundle gerbe picture, I think.

Posted by: Urs Schreiber on October 18, 2007 1:09 PM | Permalink | Reply to this

### Re: Translation into the language of Brylinski

Hi Urs,

But now lets see: does that mean that b) and c) coincide?

Yes.

Just because both involve sites/stacks etc. doesn’t mean that they are “the same”.

It does, because they are literally part of the same data. A sheaf $\mathcal{F}$ on $Man/X$ is something which assigns a set to each smooth family of points in $X$:

(1)$(U \stackrel{f}{\rightarrow} X) \mapsto \mathcal{F}(U \stackrel{f}{\rightarrow} X).$

I like to separate this definition into two pieces : what $\mathcal{F}$ does on the “single points”, and what it does on “families of points”. The former is the fundamental information, the latter is secondary. The former is (b), the latter is (c). Taken together, this is the data of a “sheaf”.

Let’s make it more explicit. What is a $G$-bundle $\mathcal{P}$ on a space $X$? We all agree that it is a smooth assignment of a $G$-torsor to every point of $X$,

(2)$x \mapsto \mathcal{P}_x.$

How do we write down “smooth assignment”? The stacky way is to say that :

An assignment $x \mapsto \mathcal{P}_x$ is smooth if you can write down a version of it which works for smooth families of points too.

In other words, in the stacky picture, a $G$-bundle is a sheaf $\mathcal{P}$ on $Man/X$, such that the fibers it assigns to points are $G$-torsors:

(3)$\mathcal{P}(pt \stackrel{x}{\rightarrow} X) \, \, is \,\, a \,\, G-torsor\,\, for\,\, each\,\, x \in X.$

That’s the “part (b)” information of the sheaf… the important one. The “part (c)” information will include the fact that the way $G$-acts must be compatible with smooth families, i.e. that $\mathcal{P}(U \stackrel{f}{\rightarrow} X)$ must carry an action of the sheaf $\overline{G}$ of functions in $G$ but that’s not important!

Let me put it differently. Say we have a conventional top-down $G$-bundle $P \rightarrow X$. Sheaf-wise, we think of $P$ as a stack $\mathcal{P}$ over $Man/X$ which assigns to a smooth family of points in $X$ the sections of $P$ over that family:

(4)$\mathcal{P}(U \stackrel{f}{\rightarrow} X) = \Gamma(f^*P) \, \, over \,\, U.$

Whenever you see a formula like this, always ask yourself : what does it do to points? Clearly it just assigns the fibers!

(5)$\mathcal{P}(pt \stackrel{x}{\rightarrow} X) = P_x.$

—-

When I wrote $\Sigma^2 U(1)$ I was thinking of a 2-group as a one-object 2-category. But I’m happy to write $\Sigma U(1)$, it looks better.

—-

Let’s face it: a “groupoid whose hom-spaces are $U(1)$-torsors” is really (a) either a groupoid enriched over $U(1)$-Tor or (b) the graph of a pseudofunctor from a codiscrete groupoid to $\Sigma U(1)$-Tor.

I’m happy with (a)… but I would prefer to keep thinking of it simply as a groupoid whose hom-sets are $U(1)$-torsors. When asked to spell that out, one quotes (a).

I’m not happy with (b) at all though. This is like saying that “a group is the graph of a functor from the codiscrete category on one object to the category of groups”. Yes, it’s true… but it’s completely over-the-top. It’s true that groupoids enriched in $U(1)-Tor$ are secretly 2-groupoids… but to me, that’s not a very important “secretly”. I might well be wrong though.

Posted by: Bruce Bartlett on October 18, 2007 2:17 PM | Permalink | Reply to this
Read the post On weak Cokernels for 2-Groups
Weblog: The n-Category Café
Excerpt: On weak cokernels of 2-groups.
Tracked: October 17, 2007 11:09 PM

### Re: What is the Fiber?

Progress is slow, but I keep building my Wiki-substitute

Now, when you go to section

Parallel $n$-Transport

subsection

Bundle gerbes

you can find some facts about bundle gerbes and how they arise canonically as descent objects for $(\Sigma U(1)- \Sigma 1d\mathrm{Vect})$-trivializable 2-transport

(as described in more detail in

and, as you guessed, in a paper in preparation with Konrad Waldorf).

Posted by: Urs Schreiber on October 18, 2007 8:44 PM | Permalink | Reply to this
Read the post n-Bundle Obstructions for Bruce
Weblog: The n-Category Café
Excerpt: On the global description of n-bundles obstructing lifts through shifted central extensions.
Tracked: November 5, 2007 9:02 AM

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