# Understanding the definition of $G$-gerbe

In Introduction to Differentiable Stacks Gregory Ginot defines a $$G$$-gerbe as the following.

Let $$G$$ be a Lie group. A $$G$$-gerbe over a stack $$\mathcal{C}$$ is a gerbe over stack $$\mathcal{D}\rightarrow \mathcal{C}$$ which locally is isomorphic to $$[pt/G]\times \mathcal{C}$$.

I am not able to understand what is locally isomorphic here.

Can some one help to clarify this.

I am trying to make sense of this by imitating what does it mean to say principal $$S^1$$ bundle over a manifold. It means a smooth map $$P\rightarrow M$$ that locally looks like product i.e., there exists an open cover $$\{U_\alpha\}$$ of $$M$$ with trivializations $$\pi^{-1}(U_\alpha)\rightarrow U_\alpha\times S^1$$. One can see this $$\pi^{-1}(U_\alpha)$$ as pull back of inclusion $$U_\alpha\rightarrow U$$ along $$\pi:P\rightarrow M$$ .

If we imitatie, by locally isomorphic we mean, I think it means there exists an atlas (open cover in above sense) $$\underline{X}\rightarrow \mathcal{C}$$ such that the fiber product $$\mathcal{D}\times_{\mathcal{C}}\underline{X}$$ (pull back $$\pi^{-1}(U_\alpha)$$ in above sense) is some how related to $$[pt/G]\times \mathcal{C}$$.

In Differentiable Stacks and Gerbes Kai Behrend and Ping Xu defines an $$S^1$$-gerbe as the following.

An $$S^1$$-gerbe over $$\mathfrak{X}$$ is a gerbe $$\mathfrak{R}\rightarrow \mathfrak{X}$$ which is locally isomorphic to $$BS^1\times \mathfrak{X}$$ and is endowed with a trivialization of its band (the $$2$$-sheeted covering $$\underline{Band}(\mathfrak{R})\rightarrow \mathfrak{X}$$).

In this also there was not much explanation of what is locally isomorphic to.

Any comments on definition of Band is welcome.

In Some notes on Differentiable stacks J. Heinloth defines a $$G$$-gerbe as the following.

A gerbe $$\mathcal{D}\rightarrow \mathcal{C}$$ is called an $$S^1$$-gerbe if there is an atlas $$\underline{X}\rightarrow \mathcal{C}$$ and a section $$s:\underline{X}\rightarrow \mathcal{D}$$ such that there is an isomorphism $$(X\times_{\mathcal{D}}X)\times_{X\times_{\mathcal{C}}X}X\cong S^1\times X$$ "as a family of groups over $$X$$" with some other conditions.

By specifying "as a family of groups on $$X$$" I think he want to see $$S^1\times X$$ as not just like a manifold but see $$S^1$$ as a Lie group and $$X$$ as a manifold separately I mean the stack associated to $$S^1\times X$$ as $$BS^1\times \underline{X}$$ and not $$\underline{S^1\times X}$$. This seems compatible with what Gregory says i.e., locally isomorphic to $$[pt/G]\times \mathcal{C}$$. Here $$G=S^1$$ and $$[pr/G]\times \mathcal{C}$$ is $$[pt/S^1]\times \mathcal{C}$$ i.e., $$BS^1\times \mathcal{C}$$.

Can some one help to clarify this.

• It should be locally equivalent, not isomorphic, since gerbes are categories. It's just an abuse of terminology though. If you can get your hands on Breen's monograph in the Asterisque series (On the classification of 2-gerbes and 2-stacks, Astérisque 225 (1994)), then there's a nice treatment of ordinary gerbes as well (and ignore the 2-gerbe stuff). Otherwise his less detailed but newer notes arxiv.org/abs/math/0611317 would suffice. – David Roberts Jan 11 at 12:23
• Say $C$ and $D$ are CFG over the same base category $S$, and $D$ is also a CFG over $C$. I would interpret the condition you mention as saying: $D$ and $[pt/G]\times C$ have to be locally (over $C$, not locally over $S$) isomorphic (in the sense of equivalent as CFG) stacks. This just means there's an atlas $X\to C$ of $C$ such that the fibered products $D\times_C X$ and $([pt/G]\times C)\times_C X = [pt/G]\times X$ are isomorphic as stacks (i.e. equivalent as CFG) over $X$ (Which is the same -I'd say but I'm not sure- as saying they are just isomorphic as stacks over $S$). – Qfwfq Jan 11 at 13:24
• @Qfwfq Oh.. ok ok. I am happy that what you are saying intersects non triviallly with what I said (existence of an atlas atleast :D)... so, you are saying it is not that $\mathcal{D}\times_{\mathcal{C}}X$ is same as that of $[pt/G]\times \mathcal{C}$ but it is same as that of $[pt/G]\times \mathcal{C}\times_{\mathcal{C}}X=[pt/G]\times X$... I think I understand what you said,,, do you have any reference other than what is suggested by David Roberts above.. – Praphulla Koushik Jan 11 at 15:00
• @DavidRoberts I have notes on 1-gerbes and 2-gerbes.. I will read that.. it only discusses $G$-gerbe over a manifold $X$ and $G$ for them is not just a group but a bundle of groups over $X$... I will read that section and ask here (after spending some 15 hours) if I did not understand something.. Thank you. – Praphulla Koushik Jan 11 at 15:11
• @Qfwfq i was just trying to imitate open cover with atlas...I don’t know what is good enough here,, atlas or representable or any random map of stacks from $X$... I will think... I will see that stacks project also... – Praphulla Koushik Jan 11 at 16:27

In Brauer Groups and Quotient stacks, they define $$G$$-gerbe as follows.

Set up : Fix a Noeth. scheme $$X$$. Let $$G$$ be a group scheme (flat, separated and of finite type) over $$X$$.

A $$G$$-gerbe over $$X$$ is a morphism $$F\rightarrow X$$, with $$F$$ an algebraic stack, such that there exists a faithfully flat map, locally of finite presentation, $$X'\rightarrow X$$ such that $$F\times_XX'\cong BG\times_XX'$$.

As $$G$$ is a group scheme over $$X$$, there is an obvious morphism $$BG\rightarrow X$$. So, we can then talk about the product $$BG\times_XX'$$. This may not be the case in case of differential geometric set up.

A morphism of stacks $$F:\mathcal{D}\rightarrow \mathcal{C}$$ is a $$G$$-gerbe, If it is a gerbe over stack in the usual sense and, I think what they mean is, there exists an atlas $$p:\underline{X}\rightarrow \mathcal{C}$$ such that the pullback $$\underline{X}\times_{\mathcal{C}}\mathcal{D}$$ is isomorphic to the stack $$BG\times_{\text{Man}}\underline{X}$$.

I also think there should be something like an action map $$BG\times_{\text {Man}}\mathcal{D}\rightarrow \mathcal{D}$$ which should be behaving like Lie group action on a manifold $$P\times G\rightarrow P$$ in case of Principal $$G$$ bundle.

• BG is not grouplike if G is not abelian, so won't act like you suggest in that case. It's more like a fibre bundle than a principal bundle... – David Roberts Jul 13 at 22:08
• I am ok if it is like a fiber bundle than a principal bundle where there is no expected action of fibere on top space.... But, I am not able to relate the justificication.. Usually, when we start with a Lie group, we talk about principal group bundles, which comes with an action, I was expecting similar thing here, something like a principal G bundle, you are saying we are not expecting a principal bundle like behaviour but we are expecting a fiber bundle like behaviour... @DavidRoberts – Praphulla Koushik Jul 13 at 22:19
• Excluding the action map, should the rest be taken as definition for G-gerbe? @DavidRoberts – Praphulla Koushik Jul 14 at 10:48
• The definition of gerbe is what it is, and a $G$-gerbe is one where all the automorphism groups of the objects are abstractly isomorphic to $G$. Together with the definition of gerbe, this implies 'locally equivalent to the classifying stack of $G$'. – David Roberts Jul 14 at 11:19
• Yes, that is right. – David Roberts Jul 14 at 11:35