## December 18, 2006

### Universal Gerbes

#### Posted by Urs Schreiber Jim Stasheff asked me to forward the following question to the $n$-Café audience:

There is a universal principal bundle for any group. Is there also a universal gerbe?

I don’t know much about the answer. The only thing I am aware of is a little trick:

$U(1)$-gerbes on $X$ are classified (depending on your taste almost by definition) by $H^3(X,\mathbb{Z})$, i.e. by homotopy classes of maps from $X$ into an Eilenberg-MacLane space $K(\mathbb{Z},3)$.

But since the group $PU(H)$ of projective unitary operators on a separable Hilbert space $H$ is a $K(\mathbb{Z},2)$, this is also the classification of $PU(H)$-bundles.

For that reason, many people who find need of $U(1)$-gerbes in their daily work tend to resort to working with $PU(H)$-bundles.

Any $PU(H)$-bundle implcitily defines a $U(1)$-gerbe: its lifting gerbe.

Constructively, given a $PU(H)$-valued transition function describing a locally trivialized $PU(H)$-bundle, we may replace the function’s values $g_{ij}(x)$ everywhere by a chosen lift $\hat g_{ij}(x)$ in the projection

(1)$U(H) \stackrel{p}{\to} PU(H)$

coming from the central extension

(2)$1 \to U(1) \to U(H) \to PU(H) \to 1 \,,$

i.e. such that

(3)$p(\hat g_{ij}(x)) = g_{ij}(x) \,.$

In general, the resulting $\hat g_{ij}(x)$ will fail to satisfy the cocycle condition

(4)$\hat g_{ij}(x) \hat g_{jk}(x) = \hat g_{ik}(x)$

but the failure is measured by a function $f$

(5)$\hat g_{ij}(x) \hat g_{jk}(x) = f_{ijk}(x) \hat g_{ik}(x) \,.$

Applying $p$ to this equation shows that $f$ takes values in the kernel of $p$, hence in $U(1)$.

While there may be no way to find a lift such that this $f$ vanishes, the $f$s will always satisfy a cocycle condition of their own, on quadruple overlaps.

This 2-cocycle charcterizes (the local trivialization of) a $U(1)$-gerbe. This is called the lifting gerbe associated to our original bundle. Its non-triviality measures the impossibility of lifting the structure group of our bundle from $PU(H)$ to $U(H)$.

For every $U(1)$-gerbe there is a $PU(H)$-bundle whose lifting to a $U(H)$-bundle is obstructed by that gerbe; $U(1)$-gerbes and $PU(H)$-bundles have the same classification.

So, given all that, we could pull a trick and declare that

The universal $U(1)$-gerbe is the lifting gerbe of the universal $PU(H)$-bundle.

That’s, at least, essentially the point of view adopted in

Alan L. Carey, Jouko Mickelsson
The universal gerbe, Dixmier-Douady class, and gauge theory
hep-th/0107207 .

For technical reasons, these authors in this article don’t quite use $U(H) \to PU(H)$, but something very similar.

Posted at December 18, 2006 2:26 PM UTC

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### Re: Universal Gerbes

The answer is almost certainly “yes”. But what was the question? I forget what you mean by the word “gerbe” Urs. I tend to mean whatever geometric gadget is convenient for the job in hand and which is classified by the third integral cohomology group (or the appropriate Deligne cohomology group if I want a connection).

Posted by: Simon on December 18, 2006 4:18 PM | Permalink | Reply to this

### Re: Universal Gerbes

The answer is almost certainly “yes”.

I also assume that the thing should exist. But I am not sure that I have seen it discussed in generality. But possibly that’s quite my fault.

But what was the question? I forget what you mean by the word “gerbe” Urs.

In this case I transmitted Jim Stasheff’s question verbatim. But I assume that an answer to the question with “gerbe” replaced by whatever equivalent structure you like would be good.

Just recently we had the classifying spaces for 2-bundles mentioned here. The question is asking for the universal something on this space, I assume, such that the thing classified by a map into it is the pullback of that something along that map.

Posted by: urs on December 18, 2006 4:57 PM | Permalink | Reply to this

### Re: Universal Gerbes

I don’t know if there’s a space, but there’s almost certainly a moduli 2-stack of G-gerbes. The most obvious functor is quite clearly a 2-stack, and it shouldn’t take too much more work (e.g. use Lurie’s Artin Approximation Theorem) to show that this moduli functor is actually an algebraic stack.

Posted by: A.J. on December 18, 2006 7:54 PM | Permalink | Reply to this

### Re: Universal Gerbes

there’s almost certainly a moduli 2-stack of $G$-gerbes

Just to ensure that we are thinking of the same something: you want to consider the 2-stack that assigns to each $X$ the bigroupoid of $G$-gerbes on $X$, right?

While I certainly agree that this should exist, I am not sure yet I understand how it would be the answer to the question. Please help me.

Maybe, let’s consider it one level down: we know what the classifying space for $G$ bundles is (namely $B G$) and we know the universal $G$-bundle on that space (namely $E G$). Plus, we know that there is a 1-stack (namely $\mathrm{Bun}_G$) of $G$-bundles (assigning $G$-bundles on $X$ to $X$). Now, in which sense does $\mathrm{Bun}_G$ tell us about $E G$?

Posted by: urs on December 18, 2006 8:05 PM | Permalink | Reply to this

### Re: Universal Gerbes

Yes, we’re thinking about the same functor. The point of introducing these moduli functors is that they provide a tautological solution to the problem. The mantra is: Any moduli stack of thingamajiggies comes naturally equipped with a universal thingamajiggy, of which any other thingamajiggy is a pullback.

So construct a moduli stack of gerbes, and you get, by nonsense alone, a universal gerbe. For instance, the moduli space M of curves has a natural curve over it, whose fiber over any point C in M is just the curve C. Likewise, the product of Bun_G and C has a universal bundle, whose fiber over the point (P,c) is just the torsor P_c.

Posted by: A.J. on December 19, 2006 8:02 PM | Permalink | Reply to this

### Re: Universal Gerbes

The answer to Jim Stasheff’s question is surely yes; the problem is that there are various ways to formulate this question and prove the answer is yes. Given a topological group $G$, we can form the topological 2-group $AUT(G)$, and then form the geometric realization of the nerve of this, say $|AUT(G)|$, which is a new topological group. A $G$-gerbe over a space $X$ should be ‘morally equivalent’ to an $|AUT(G)|$-bundle over $X$, at least if $X$ is a nice space (paracompact or something).

If this is indeed true, the space $B(|AUT(G)|)$ should in some sense be the classifying space for $G$-gerbes. In particular, there should be a universal $G$-gerbe over this space such that any $G$-gerbe over $X$ is equivalent to the pullback of this one along some map $f: X \to B(|AUT(G)|)$.

The problem is making all this precise. I believe Danny Stevenson will have something to say about all this in his soon-forthcoming paper - not necessarily for $AUT(G)$, but maybe for more general or possibly just different topological 2-groups (like the string 2-group).

None of the mistakes I might be making here are his fault, of course.

Posted by: John Baez on December 18, 2006 8:31 PM | Permalink | Reply to this

### Re: Universal Gerbes

[…] morally equivalent […]

I should have mentioned this, since it is the generalization to general 2-groups of what, as described above, Carey and Mickelsson do for the special 2-group $G_2 = \Sigma(\Sigma(U(1)))$.

We discussed all this a while ago here, where also some (but not all) details can be found for how to describe a $G_2$-3-cocycle as a $|G_2|$-2-cocycle.

Posted by: urs on December 18, 2006 8:43 PM | Permalink | Reply to this

### Re: Universal Gerbes

Urs wrote:

We discussed all this a while ago here, where also some (but not all) details can be found for how to describe a $G_2$-3-cocycle as a $|G_2|$-2-cocycle.

I believe Danny is trying to fill in all these details in an elegant way in his forthcoming paper. He plans to run this paper by me over the Christmas break. After that I’ll have a much better sense of what he’s done — but also, he’ll put his paper on the arXiv.

Posted by: John Baez on December 18, 2006 9:40 PM | Permalink | Reply to this

### Re: Universal Gerbes

[…] fill in all these details […]

I am looking forward to seeing that. Danny kindly promised to explain it to me in Toronto.

As far as I recall, back then we were able to write down proofs that showed that, for $G_2$ any strict 2-group, any $G_2$-3-cocycle gives rise to a $|G_2|$-2-cocycle.

(There are two different conventions for numbering these cocycles. If you think the above statement looks odd, just lower every cocycle index by one.)

Danny did that, as far as I remember, in terms of crossed modules. I did write down what should be the “2-anafunctor” # version of this proof #, where we consider an $n$-cocycle as an ($n-1$)-functor on a certain $(n-1)$-categorical surjection of the discrete $(n-1)$-category over base space.

As far as I remember, the question was if and how this $H^3(X,G_2) \to H^2(X,|G_2|)$ has an inverse.

I remember that Larry Breen kindly looked into this when we met in Vienna, indicating that this should obviously be true, but that Danny kept pointing out some problem, at which point, it seems, I lost track of what was going on.

Posted by: urs on December 19, 2006 9:16 AM | Permalink | Reply to this

### Re: Universal Gerbes

I wrote:

We discussed all this a while ago here,

As I point out there, but should mention here again: this discussion was to a large part induced by theorem 5.6 in

(p. 12), which states essentially what John Baez recalled above #, namely that (generalizing from automorphism 2-groups to general strict 2-groups), $(H \to G)$-gerbes have the same calssification as $|(H \to G)|$-bundles.

(Here $(H\to G)$ denotes a (topological) strict 2-group coming from a crossed module of the two ordinary groups $G$ and $H$, and $|H\to G|$ denotes its nerve.)

As far as I understood, Danny Stevenson’s work, mentioned above, is mainly supposed to clarify one direction of the proof of this.

Posted by: urs on December 20, 2006 10:10 AM | Permalink | Reply to this

### Re: Universal Gerbes

I have a problem with the physics underlying the Mickelsson-Carey construction in section 1.1. To each connection A they associate a fermionic Fock space VA. The group of unitaries only acts projectively on VA, which gives rise to an obstruction in H3(G,Z); hence the relation to gerbes.

My problem with this construction is that it couples quantum chiral fermions to a classical background gauge field. This is fine for understanding anomalies, but fundamentally there is no classical gauge field; it should be quantized as well. But if you try to quantize the gauge field, it can no longer be used to define a polarization, and Pickrell’s no-go theorem comes back to haunt you.

For a more low-brow discussion (appropriate for me) of Fock spaces parametrized by connections and their relation to chiral anomalies, try
Luis Alvarez-Gaumé and Philip Nelson
Hamiltonian interpretation of anomalies
Comm. Math. Phys. 99, no. 1 (1985), 103–114