The String Coffee Table

Consider the 2-group $\Sigma (U(1))$: a single object, $U(1)$ worth of morphisms.

Try to build a bundle of categories whose typical fiber is a torsor for that.

Since $\Sigma (U(1))$ is strict, you’d guess strict 2-torsors would do. But they don’t.

So weaken things. As a category in sets, $\Sigma (U(1)$ is equivalent to the 2-group $\mathrm{Tor}(U(1))$, whose objects are $U(1)$-torsors and whose morphisms are torsor morphisms (HDA V, p. 51).

A $U(1)$-torsor is a circle $S^1$. We need to take care not to identitfy different circles $S^1$, ${\tilde{S}}^1$, etc.

Hence $\mathrm{Tor}(U(1))$ contains a lot of objects. How many circles are there, in the world?

But what we want is a topological 2-torsor. So we want not just all circles, but all continuous families of them.

In other words, what we are really looking for is not the collection of all circles, but the moduli space of circles.

Pick any topological space $X$ and consider the collection of continuous families of circles parameterized by $X$. This collection is the same as maps from $X$ into the moduli space of circles, otherwise known as $\mathrm{BU}(1)$, the classifying space for circle bundles.

Hence we want our $\Sigma (U(1))$-torsor to have a space of objects which looks like $\mathrm{BU}(1)$.

In fact, each point in this space corresponds to a circle, in the sense that we have the canonical circle bundle on $\mathrm{BU}(1)$, known as the universal princiapl $U(1)$-bundle $\mathrm{EU}(1)$.

Given two circles $S^1$ and ${\tilde{S}}^1$, the space of torsor morphisms between them is

Similarly for any continuous family of circles.

Hence we find that the space of morphisms of our $\Sigma (U(1))$-torsor is

with the obvious source and target maps

In other words, $T$ is nothing but the transport groupoid of the universal $U(1)$-bundle

Let’s try to build a bundle of categories with typical fiber $T$.

Take any $\mathrm{BU}(1)$-bundle $E\to X$, classified by an element in $H^3(X,\mathbb{Z})$. Assume we can find a continuous map from the total space $E$ to $\mathrm{BU}(1)$ which restricts on each fiber to a representative of the canonical circle bundle.

We get a category internal to the category of bundles

whose bundle of objects is $E$ and whose bundle of morphisms is $\frac{L{\times }_{X}L}{U(1)}$, where $L$ is, on each fiber of $E$, the canonical circle bundle.

The same game can be played for more general 2-groups. Let $(H\to G)$ be any crossed module, regarded as a strict 2-group.

Again, regarded in sets this is equivalent to ${\mathrm{Bitor}}_G(H)$, which are $H$ bitorsors whose left and right action are related by an element in $G$ (just ordinary bitorsors if $G=\mathrm{Aut}(H)$).

Again, this is huge. How many $H$ bitorsors are there in the world?

What we want is continuous families of them. As before, this are nothing but $H$-bitorsor bundles whose left and right actions are related by elements in $G$.

The moduli space of all $(G-H)$-bitorsors is known. It’s the classifying space for $(G-H)$-bibundles, known as the geometric realization of the nerve $\mid (H\to G)\mid$.

As before, we would want to build bundles of categories with typical fiber the transport groupoid of the universal $H$-bibundle over $\mid (H\to G)\mid$.

Say $(H\to G)$ is the string 2-group ($\to$), such that $\mid (H\to G)\mid$ is $\mathrm{String}$ itself. We’d start with a $\mathrm{String}$-bundle and re-interpret each fiber as a category, namely as the transport groupoid of the universal $(G-H)$-bibundle.

Up to the fact that I don’t know if this the construction I am describing always exists as a continuous thing.