MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

So a $2$-group presented by a crossed module

$H\overset{t}{\to} G$,

where $t$ has nontrivial kernel and cokernel, is given just by the data of that kernel and cokernel, the action of the cokernel on the kernel, and the class in the group cohomology $H^3(\text{coker } t, \text{ker } t)$ classifying the extension above. Therefore, a principal $2$-bundle with fiber the $2$-group above (or at least a fake-flat one (?)) should also be presented by this data.

So, how do you do it?

I think that the extension class can be interpreted as a $2$-bundle with fiber $\text{ker t}$ associated via the action I mentioned to some principal $2$-bundle with fiber $\text{coker }t$ over the classifying space $B\text{coker t}$. Then the classifying map of the residual $\text{coker }t$ bundle on my space can be used to pull back this 2-bundle. Is its topological class the same as the principal $2$-bundle with fiber $\text{ker }t$ from the reduction of structure group of my original bundle (assuming it's fake-flat)?


share|cite|improve this question
up vote 1 down vote accepted

Here is the general story (as in section 4.3 of Principal infinity-bundles -- General theory).

So consider

$$ A \stackrel{i}{\to} \hat G \stackrel{p}{\to} G $$

a "central" extension of higher groups, hence a long homotopy fiber sequence of the form

$$ \array{ \mathbf{B}A &\stackrel{}{\to}& \mathbf{B}\hat G \\ && \downarrow \\ && \mathbf{B}G &\stackrel{\mathbf{c}}{\to}& \mathbf{B}^2 A } \,, $$

where the map $\mathbf{c}$ is the cocycle that classifies the extension. For instance if $A = \mathbf{B} ker(t)$ then this is a higher group 3-cocycle on $G$ with coefficients in $ker(t)$.

Now by the pasting law for homotopy pullbacks, if you have a $\hat G$-principal infinity-bundle modulated by a map

$$ X \stackrel{\hat g}{\to} \mathbf{B}\hat G $$

with, hence, underlying $G$-principal $\infty$-bundle modulated by

$$ g : X \stackrel{\hat g}{\to} \mathbf{B}\hat G \stackrel{}{\to} \mathbf{B}G $$

(this may be an ordinary principal bundle if $G$ is a 0-truncated $\infty$-group, as in your case)

then we get the pasting diagram of homotopy pullbacks of the form

$$ \array{ \hat G &\to& \hat P &\to& \ast \\ \downarrow && \downarrow && \downarrow \\ G &\to& P &\to& \mathbf{B}A &\to& \ast \\ \downarrow && \downarrow && \downarrow && \downarrow \\ \ast &\stackrel{x}{\to}& X &\stackrel{\hat g}{\to}& \mathbf{B}\hat G &\stackrel{}{\to}& \mathbf{B}G } \,. $$

Here we read off

  • $\hat P \to X$ is the $\hat G$-principal $\infty$-bundle modulated by $\hat g$

  • $P \to X$ is the underlying $G$-principal $\infty$-bundle;

  • $\hat P \to P$ is an $A$-principal $\infty$-bundle on the total space of $P$ with the special property that restricted to the fibers it becomes the extension $\hat G \to G$ that we started with

This is the stage-wise decomposition of principal $\infty$-bundles which is induced from the extension of $\infty$-groups that we started with. This construction establishes an equivalence of $\infty$-categories between $\hat G$-principal $\infty$-bundles on $X$ and $A$-principal $\infty$-bundles on total spaces of $G$-principal $\infty$-bundles satisfying these compatibility conditions.

An important example of this is the case where $\hat G$ is the smooth string 2-group sitting in the extension

$$ \mathbf{B} U(1) \to String \to Spin \,. $$

and classfied by the smooth refinement of the first fractional Pontryagin class $\frac{1}{2}\mathbf{p}_1$. Here the above tells us that String-2 bundles are equivalently circle 2-bundles on total spaces of Spin-principal bundles which restrict to the canonical bundle gerbe on each fiber.

(I could say more, but I need to stop as the updating of the formula typesetting almost kills my little computer now. This used to be better with the previous MO version, where I could select "one shot math preview"...)

share|cite|improve this answer
Thanks, Urs! This is a really nice answer so far. I'm still a bit confused though. Is $\hat{G}$ the same as my 2-group $H\to G$? – Ryan Thorngren Jul 6 '13 at 3:09
Nevermind. Now I see that it must be. Is the picture I described in my last paragraph the same as this description? – Ryan Thorngren Jul 6 '13 at 3:19
@RyanThorngren Huh - I think I got the wrong end of the stick when I wrote my answer. Now I read it again in conjunction with Urs' answer, I get what you mean. – David Roberts Jul 6 '13 at 6:48
Hi Ryan, yes, $\hat G$ would be the 2-group, and $G = coker(t)$ etc. But what you indicated in the last paragraph seemed to be pulling back the bundle $\mathbf{B}\hat G \to \mathbf{B}G$, along $X \to \mathbf{B}\hat G \to \mathbf{B}G$, right? One can play this way, too, then one gets maps into the Cech-nerve of this bundle. But let me know if I am missing your point. – Urs Schreiber Jul 6 '13 at 9:40
Another thing I wanted to say which then the math-formatting prevented me from is how the connections come in. One useful way to proceed is like this: Consider the smooth universal map $\mathbf{c} : \mathbf{B}G \to \mathbf{B}^2 A$ that classifies the extension $\mathbf{B}\hat G$. Suppose you know already how to differentially refine that to a differential cocycle $\mathbf{c}_{conn} : (\mathbf{B}G)_{conn} \to \mathbf{B}^2 A_{conn}$. Then one wants to define $(\mathbf{B}\hat G)_{conn}$ to be a homotopy fiber of $\mathbf{c}_{conn}$. But there are now further sensible choices: – Urs Schreiber Jul 6 '13 at 9:44

One way to think of the setup is as the groupoid underlying the 2-group being a $\operatorname{ker} t$ bundle gerbe (i.e. a 2-bundle) over $\operatorname{coker} t$ in the category of groups.

Another way to think of this is as an extension of the delooping $\mathbf{B}\operatorname{coker} t$ (a one-object groupoid) by the double delooping $\mathbf{B}^2\ker t$ (a one-object, one-arrow 2-groupoid), giving the one-object 2-groupoid $\mathbf{B}\mathcal{G}$ - here I've written $\mathcal{G}$ for the 2-group. This latter interpretation makes the connection to group cohomology clearer.

Taking the nerve and geometric realisation of this latter version gives you a topological $B^2\operatorname{ker} t$ bundle over $B\operatorname{coker}t$ (using the fact $B\operatorname{ker}t$ is an abelian group, here). One can then construct the lifting bundle gerbe for the extension $$ B\operatorname{ker}t \to EB\operatorname{ker}t \to B^2\operatorname{ker}t $$ to get a $B\operatorname{ker}t$ bundle gerbe, and then construct the lifting bundle 2-gerbe for the extension $$ \operatorname{ker}t \to E\operatorname{ker}t \to B\operatorname{ker}t, $$ giving you a $\operatorname{ker}t$-2-bundle gerbe, which is a 3-bundle, rather than a 2-bundle.

It then depends what you want, and what you want to do with it, as to which construction you use.

share|cite|improve this answer
Thanks for your answer, David. I'm beginning to understand it more. Here you are also considering the case where the cokernel acts trivially on the kernel, correct? And does the cocycle $(\text{coker }t)^3\to \text{ker }t$ describe the Cech transition maps for the $B\text{ker }t$ 2-bundle over $B\text{coker }t$ that realises the classifying space of the 2-group? – Ryan Thorngren Jul 10 '13 at 14:59
By the way, I'm starting a discuss at the nforum and you're welcome to respond there if you'd prefer. – Ryan Thorngren Jul 10 '13 at 15:08

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.