U(1) vs. BZ and representations of 2-groups

Question

$U(1)$ seems to lead a dual life. On one hand it is the group we know and love, and on the other, it is the classifying space of the integers. Thinking about $n$-groups says that we should also think about this delooping $B\mathbb{Z}$ as the 2-group presented by the crossed module $\mathbb{Z} \to 1$. What is the relationship between this and $U(1)$ as a group?

In particular, let me ask what this relationship says about projective representations of a group. That is, if we have a projective representation of a group $G$, we get a cocycle in $H^2(G,U(1))$. This defines a central extension

$1 \to U(1) \to \hat G \to G \to 1$,

to which our representation lifts to an ordinary representation. We can also think about this cocycle as living in $H^3(G,\mathbb{Z})$, which then defines a central extension of 2-groups

$1 \to B\mathbb{Z} \to \tilde G \to G \to 1$,

or equivalently a 4-term exact sequence with $\mathbb{Z}$ on the left. Can I see the projective representation in terms of $\tilde G$ or this 4-term exact sequence?

Thanks.

Answer 1

This is about the difference between and relation of "bare" groups and groupoids and higher groupoids (bare = no extra geometry) and smooth groups and groupoids and higher groupoids.

The main fact is that there is an $\infty$-functor

$$ \Pi : Smooth \infty Grpd \to \infty Grpd $$

which is $\infty$-left adjoint to the one that equips $\infty$-groupoids with their discrete smooth structure, and this is geometric realization of smooth infinity-groupoids .

This functor is a reflection: it sends discretely smooth $\infty$-groupoids such as $\mathbf{B}^n \mathbb{Z}$ to themselves, with the geometry forgotten. But otherwise it is far from being injective.

We can say that for $A \in \infty Grpd$ any bary $\infty$-groupoid, that a lift through $\Pi$, hence a smooth infinity-groupoid $\mathbf{A}$ with $\Pi(\mathbf{A}) \simeq A$, is a smooth refinement of $A$.

For example $\mathbf{B}^n U(1)$ is naturally a smooth $\infty$-groupoid for all $n \in \mathbb{N}$ and one finds that

$$ \Pi(\mathbf{B}^n U(1)) \simeq B^{n+1} \mathbb{Z} \simeq K(\mathbb{Z}, n+1) \,. $$

Here on the left the "$\mathbf{B}$" is in boldface just to amplify that this is the deloopoing in smooth $\infty$-groupoids, whereas the "$B$" on the right is the traditional delooping in bare homotopy types.

Also $\mathbf{B}^{n+1}\mathbb{Z}$ exists as a (discretely) smooth $\infty$-groupoid, but is as such not equivalent to $\mathbf{B}^n U(1)$. Nevertheless, both are smooth refinements of $B^{n+1} \mathbb{Z}$.

Another thing relevant to your question is the following: for nice enough smooth $\infty$-groups $G$ it happens that while $\mathbf{B}G$ is very different from $B G := \Pi(\mathbf{B}G)$, both classify $G$-principal $\infty$-bundles on equivalence classes . But the smooth version only sees also the right homotopy type of smooth gauge transformations of these.

This follows from the fact that for nice enough $G$ homotopy pullbacks over $\mathbf{B}G$ are preserved by $\Pi$. This also implies that in these cases $\Pi$ preserves the $\infty$-group structure.