Delooping groupoid

Question

I have seen the notion of a delooping groupoid defined as a 1-groupoid having a single object * and the automorphisms of * to be the group G, with the composition of 1-morphisms being the usual group operation. In this sense, a group is a pointed 0-connected groupoid. In higher cases too, an n-group is a pointed connected n-groupoid.

But I have seen this for a discrete group G. What about a continuous group? Say, we take G to be U(1) and then form BU(1). Is it a 1-groupoid or a 2-groupoid? I have seen claims that BU(1) is a 2-group. I guess it follows from the fact that the homotopy type of U(1) is equivalent to B$\mathbb{Z}$, making BU(1) like B$^{2}\mathbb{Z}$.

Could someone briefly explain how to see this as a 2-group? Do we have to always bring this to something relating to discrete groups? Thanks.

Answer 1

Say, we take G to be U(1) and then form BU(1). Is it a 1-groupoid or 2-groupoid?

The delooping of the Lie group $\def\B{{\sf B}} \def\U{{\sf U}} \U(1)$ is the Lie 1-groupoid $\B\U(1)$.

The shape $\def\sh{\smallint}\sh\B\U(1)$ of the Lie 1-groupoid $\B\U(1)$ can be computed as follows: $$\def\R{{\bf R}}\def\Z{{\bf Z}} \sh\B\U(1)≅\sh\B(\R/\Z)≅\B\sh(\R/\Z)≅\B(\sh\R/\sh\Z)≅\B(\B\Z)≅\B^2\Z.$$ Thus, the shape of the Lie 1-groupoid $\B\U(1)$ happens to be a 2-truncated ∞-groupoid (alias 2-groupoid), namely, $\B^2\Z$, the double delooping of integers.

I have seen BU(1) to be the 2-group.

Yes, as a Lie 1-groupoid, $\B\U(1)$ possesses a compatible monoidal structure that turns it into a Lie 2-group. The Lie 2-group $\B\U(1)$ itself has a delooping $\B^2\U(1)$, and this process can be repeated indefinitely. The shape of the Lie $(n+1)$-group $\B^n\U(1)$ is $\B^{n+1}\Z$, an $(n+1)$-truncated ∞-groupoid, in fact, an abelian $(n+2)$-group.