# Central extension gives a gerbe over stack

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.

• 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. – David Roberts Jan 13 '19 at 20:05
• @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? – Praphulla Koushik Jan 13 '19 at 20:12
• 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. – David Roberts Jan 13 '19 at 20:24
• @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? – 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.