Consider a central extension of Lie groups $1\rightarrow S^1\rightarrow \hat{G}\xrightarrow{\pi} G\rightarrow 1$.

I understand that this mean $\pi:\hat{G}\rightarrow G$ is a surjective homomorphism of Lie groups (not sure if this has to be submersion) and that $S^1\subseteq Z(\hat{G})$. There is a local section $\sigma:U\rightarrow \hat{G}$ such that $\pi\circ \sigma=1_U$ where $U$ is an open nbd of $1\in G$. Correct me if I am missing some conditions.

Let $X$ be a manifold with an action of $G$ on it. Then we have the notion of quotient stack $[X/G]$.

There is an action of $\hat{G}$ on $X$ given by $(\hat{g},x)\mapsto \pi(\hat{g})\cdot x$.

We can then consider the quotient stack $[X/\hat{G}]$.

Given a manifold $Y$, objects of $[X/G](Y)$ are pairs $(P\rightarrow Y,P\rightarrow X)$ where $P\rightarrow Y$ is a principal $G$ bundle and $P\rightarrow X$ is a $G$-equivariant space (see that $G$ acts on $P$ and $X$).

"As locally any map $T\rightarrow G$ can be lifter to $\tilde{G}$", the map of stacks $[X/\hat{G}]\rightarrow [X/G]$ is a gerbe over stack.

I see that, locally any map $\theta: T\rightarrow G$ can be lifted to $\hat{G}$. As there is a section $\sigma:U\rightarrow \hat{G}$, we can consider $\theta^{-1}(U)\xrightarrow{\theta} U\xrightarrow{\sigma} \hat{G}$ and $\pi\circ (\sigma\circ \theta)=\theta$. Thus, any map $\theta:T\rightarrow G$ can be locally lifted to $\hat{G}$. But, I am not able to see why this imply $[X/\tilde{G}]\rightarrow [X/G]$ is a gerbe over stack.

Any comments are welcome.

  • $\begingroup$ You can consider the special case of $X=pt$, since the example you give is pulled back along $[X/G] \to [pt/G]$, and the pullback of a gerbe is a gerbe. $\endgroup$ – David Roberts Jan 13 '19 at 20:05
  • $\begingroup$ @DavidRoberts I am getting confused with English ... You are asking to take $X=pt$ and then saying consider $[X/G]\rightarrow [pt/G]$... Are you saying consider the obvious map of stacks $[X/G]\rightarrow [*/G]$ and pull back $[*/\hat{G}]\rightarrow [*/G]$ along $[X/G]\rightarrow [*/G]$ to get $[X/\hat{G}]\rightarrow [X/G]$?? As $[*/\hat{G}]\rightarrow [*/G]$ is a gerbe over stack, so is the pull back $[X/\hat{G}]\rightarrow [X/G]$?? Is this what you mean? $\endgroup$ – Praphulla Koushik Jan 13 '19 at 20:12
  • $\begingroup$ Then, also, it should be easier to see why the stack $[pt/\hat{G}]$ of principal $\hat{G}$-bundles is a gerbe over $[pt/G]$. Given any $X\to [pt/G]$, that is, a principal $G$-bundle $P\to X$, there is a cover $U\to X$ such that $U\to X \to [pt/G]$ lifts to $[pt/\hat{G}]$: just take a trivialising cover for $P$. Thus $[pt/\pi]$ is an epimorphism of stacks. A similar type of thinking—unwinding the definition of the stack in terms of bundles—will help to show that $[pt/\hat{G}] \to [pt/\hat{G}] \times_{[pt/G]} [pt/\hat{G}]$ is also an epimorphism. $\endgroup$ – David Roberts Jan 13 '19 at 20:24
  • $\begingroup$ In response to your comment: yes. $\endgroup$ – David Roberts Jan 13 '19 at 20:25
  • $\begingroup$ @DavidRoberts Ok. Thanks for the clarification :) First I have to prove that $[*/\hat{G}]\rightarrow [*/G]$ is a gerbe over stack and then prove that pull back (fiber product) of gerbe over stack is a gerbe over stack (it may be obvious but I did not prove yet).. It looks like it has nothing to do with central extension... Any morphism of Lie groups $\hat{G}\rightarrow G$ such that there is a local section $U\rightarrow \hat{G}$ should give a gerbe over stack $[*/\hat{G}]\rightarrow [*/G]$... Is that the case? $\endgroup$ – Praphulla Koushik Jan 13 '19 at 20:30

As discussed in the comments, $[X/\hat{G}] \to [X/G]$ is the pullback of $[pt/\hat{G}] \to [pt/G]$ along the canonical map $[X/G] \to [pt/G]$, so it suffices to show that $[pt/\hat{G}] \to [pt/G]$ is a gerbe. Since every principal $G$-bundle is locally trivial, it can be locally lifted to a principal $\hat{G}$-bundle, which is another way of saying that $[pt/\hat{G}] \to [pt/G]$ is an epimorphism of stacks. Using the fact that $\hat{G}\to G$ is surjective, then there is an equivalence of stacks $$ [pt/\hat{G}\times_G\hat{G}] \stackrel{\simeq}{\to} [pt/\hat{G}]\times_{[pt/G]}[pt/\hat{G}] $$ Thus the diagonal $[pt/\hat{G}] \to [pt/\hat{G}]\times_{[pt/G]}[pt/\hat{G}]$ is equivalent to $[pt/\hat{G}] \to [pt/\hat{G}\times_G\hat{G}]$, induced by the diagonal homomorphism $\hat{G} \to \hat{G}\times_G\hat{G}$. Such a map of stacks is an epimorphism by the same argument as before, and so we are done.

  • 1
    $\begingroup$ You mean “which is another way of saying” when you said “which is not another way of saying “ $\endgroup$ – Praphulla Koushik Feb 21 '19 at 11:17

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