# Geometric models for 2-gerbes

One can think of a complex line bundle as a geometric model for an integral cohomology class of degree 2. Similarly, a locally-trivial bundle of $$C^*$$-algebras with fiber B(H) (the $$C^*$$-algebra of bounded operators on an infinite-dimensional Hilbert space) can be thought of as a geometric model for an integral cohomology class of degree 3. One can say that such a bundle is a 1-gerbe, while a complex line bundle is a 0-gerbe. Are there similarly nice models for integral cohomology classes of degree 4 (that is, for 2-gerbes)?

• $E_8$ principal bundles? :) – Aaron Bergman Jan 14 at 18:25
• One silly answer is: your description of 0 and 1 gerbes is equivalent to the observation that $K(\mathbb{Z},2)=BU(1)$ and $K(\mathbb{Z},3)=BPU(H)$. So we want to identify $K(\mathbb{Z},4)$ as $BG$ for some topological group $G$. But of course we could take $G=K(\mathbb{Z},3)$ and find that $2$-gerbes are principal $K(\mathbb{Z},3)$ bundles... – John Greenwood Jan 14 at 18:41
• It might help to understand what it is you’re looking to do. A four form is just a map to a $K(Z,4)$, and that is fairly geometric. As John said, that’s the same as a $K(Z,3)$-bundle, which you can model as $E_8$ in reasonably low dimensions or use some other example of a $K(Z,3)$ with a group structure. I think you can even work directly with bundles of homotopy types in $\infty$-land. You could also model it as a differential cohomology class. Or you can look up the definition of a 2-gerbe in Breen’s work. I’m not sure any of these is better or worse than any of the others. – Aaron Bergman Jan 16 at 2:50
• One requirement is that the model is geometric, i.e. does not involve objects which are only defined up to homotopy equivalence. Modeling a degree-$n$ cohomology class as a map to $K(Z,n)$ is not geometric, because $K(Z,n)$ is a homotopy type, not a concrete space. Similarly, modeling a degree-2 class as a map to $BU(1)$ is not geometric, because the classifying space of a group is defined only up to homotopy equivalence. A further more vague requirement is that I want a model which has a physics flavor. Roughly, it should involve Hilbert spaces in in some way. – Anton Kapustin Jan 17 at 16:55

One fairly concrete way to view these things is via the Cech model (= transition functions). But maybe this isn't what you were looking for...

Say the base space $$X$$ is a manifold or finite $$CW$$-complex. Choose a cover by contractible open sets $$\amalg U_{i}\rightarrow X$$.

A line bundle on $$X$$ is the data of functions $$f_{ij}:U_{ij}\rightarrow U(1)$$ on double overlaps, satisfying certain conditions on triple overlaps $$U_{ijk}$$.

A 1-gerble on $$X$$ is the data of line bundles $$L_{ij}\rightarrow U_{ij}$$ on double overlaps, isomorphisms $$\phi_{ijk}:L_{ij}\otimes L_{jk}\rightarrow L_{ik}$$ on triple overlaps, satisfying certain conditions on quadruple overlaps $$U_{ijkl}$$. Assuming the cover is nice enough, this is equivalent to functions $$f_{ijk}:U_{ijk}\rightarrow U(1)$$ satisfying conditions on $$U_{ijkl}$$.

So a 2-gerble would be either: assign a 1-gerble to each double overlap, as well as the appropriate coherence data on higher overlaps (which seems baffling to me), or: repeat the 1-gerbe recipe but start at triple overlaps, or: repeat the 0-gerbe recipe but start at quadruple overlaps.

• I guess I should have made clear that I want a description which does not involve a choice of a cover. The descriptions of 0-gerbes and 1-gerbes that I gave satisfy this requirement. – Anton Kapustin Jan 14 at 21:32

Two possible ideas. One is that you can realize $$K(\mathbb{Z},3)$$ as the quotient $$U(HS)/PU(\infty)$$ where $$U(HS)$$ is the unitary group on the Hilbert space of Hilbert-Schmidt operators. However, I don’t know if you can see the group structure in this model to make a principal bundle. This construction is in https://arxiv.org/pdf/hep-th/9702147.pdf.

There is also Andre Henriques’s conjecture that the outer automorphisms of a hyperfinite type III factor model $$K(\mathbb{Z},3)$$. See `Naturally occuring' $K(\pi, n)$ spaces, for $n \geq 2$.

• The second option sounds good to me! I need to watch this lecture again (I already did once, years ago, but forgot most of it). – Anton Kapustin Jan 19 at 15:40