One can think of a complex line bundle as a geometric model for an integral cohomology class of degree 2. Similarly, a locallytrivial bundle of $C^*$algebras with fiber B(H) (the $C^*$algebra of bounded operators on an infinitedimensional 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 1gerbe, while a complex line bundle is a 0gerbe. Are there similarly nice models for integral cohomology classes of degree 4 (that is, for 2gerbes)?

1$\begingroup$ $E_8$ principal bundles? :) $\endgroup$ – Aaron Bergman Jan 14 at 18:25

1$\begingroup$ 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... $\endgroup$ – John Greenwood Jan 14 at 18:41

$\begingroup$ 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 2gerbe in Breen’s work. I’m not sure any of these is better or worse than any of the others. $\endgroup$ – Aaron Bergman Jan 16 at 2:50

$\begingroup$ 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 degree2 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. $\endgroup$ – 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 1gerble 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 2gerble would be either: assign a 1gerble to each double overlap, as well as the appropriate coherence data on higher overlaps (which seems baffling to me), or: repeat the 1gerbe recipe but start at triple overlaps, or: repeat the 0gerbe recipe but start at quadruple overlaps.

2$\begingroup$ I guess I should have made clear that I want a description which does not involve a choice of a cover. The descriptions of 0gerbes and 1gerbes that I gave satisfy this requirement. $\endgroup$ – 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 HilbertSchmidt 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/hepth/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 $n \geq 2$.

$\begingroup$ 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). $\endgroup$ – Anton Kapustin Jan 19 at 15:40