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 '20 at 18:25

2$\begingroup$ 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 '20 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 '20 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 '20 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 '20 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
 Alan L. Carey, Diarmuid Crowley, Michael K. Murray, Principal Bundles and the Dixmier Douady Class, Commun.Math.Phys. 193 (1998) pp 171196, doi:10.1007/s002200050323, arXiv:hepth/9702147.
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 '20 at 15:40
Bundle $n$gerbes seem to be what you are looking for.
Bundle gerbes can be defined w.r.t. an open cover (then it is what John Greenwood wrote), but don't have to. Instead of a cover, any surjective submersion is ok. $k$fold intersections are then replaced by $k$fold fibre products.
A nice example are basic bundle gerbes over compact Lie groups. For the groups $SU(n)$ and $Sp(n)$, one can construct them using an open cover of the Lie group, with a canonical line bundle over the double intersections. For the other simplyconnected Lie groups, one needs to "resolve" the open sets by principal bundles over them, related to certain stabilizer subgroups. The union of their total space then gives a surjective submersion mapping to $G$, over whose 2fold fibre product then again a canonical line bundle can be defined.
A further example are socalled lifting bundle gerbes, which are constructed from the problem of lifting the structure group of a principal bundle along a central extension. In this case, the surjective submersion is the bundle projection.
For bundle 2gerbes, there is the ChernSimons bundle 2gerbe. Its surjective submersions is the bundle projection of a principal $G$bundle. Over its 2fold fibre product, it has a bundle gerbe, over its threefold fibre product, a bundle gerbe isomorphism, over its 4fold fibre product, a 2isomorphism, and over its 5fold fibre product, a condition. The ChernSimons bundle 2gerbe represents a class in $H^4(M,\mathbb{Z})$, the level of a ChernSimons theory over $M$. It also carries a canonical connection, whose 3holonomy is the value of the ChernSimons theory on closed oriented 3manifolds. This works for arbitrary Lie groups $G$, with no need to assume simplyconnectedness.

$\begingroup$ I think bundle 2gerbes will work for my purposes, thanks. $\endgroup$ – Anton Kapustin Feb 1 at 16:31