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Does there exist a finite (abstract) group $G$ and a non-trivial $G$-gerbe $\mathcal X\to \mathbb C$, where we work in the category of analytic stacks.

My guess is that $G$-gerbes for $G$ an abelian group are classified as usual by $H^2_{an}(\mathbb C,G)$ and that this group is trivial.

What about non-abelian groups?

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In fact you have the following general result: Let $G$ be a Lie group, $M$ a manifold, and $p:P\rightarrow M$ a gerbe such that for every open subset $U$ of $M$, the objects of the fibre of $U$ are principal $G$-bundles. Suppose that the following condition (T) is satified:

There exists a covering $(U_i)_{i\in I}$ of $M$ such that for every pairwise distinct $i,j,k,l\in I$, $U_i\cap U_j\cap U_k\cap U_l$ is empty and the fibre of $U_i$ is not empty, then the gerbe is trivial i.e it has a global section since you can use refinement of the $(U_i)_{i\in I}$ to write the non commutative classifying cocycle whose boundary is obviously trivial here.

The same is true for every finite group $G$, You can assume that the objects of $\chi$ are $G$-principal bundles defined over open subsets of $C$. Let $B(0,r)$ be an open ball whose fibre is not empty. Suppose that $r$ is maximal for these properties and $r<+\infty$. For every $x$ in the boundary of $B(0,r)$ there exists $B(x,r_x)$ such that the restriction of the gerbe on $B(x,r_x)$ has a section $e_x$ , you can cover the circle $C_r$ with a finite number of $B(x_{x_i},r_{x_i})=B(x_i,r_i) i=1,...,n$ such that the fibre of $B(x_i,r_i)$ is not trivial. You can suppose that $B(0,r),B(x_i,r_i)$ $i=1,...n$ verify the condition (T). Thus the gerbe has a section on $B(0,r)\cup_{i=1,..,n} B(x_i,r_i)$ which contains $B(0,r'), r'>r$. This is a contradiction with the fact that $B(0,r)$ is a maximal ball whose fibre is not empty.

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