# Fiberwise criterion for a stack to be a gerbe

Let $f:X\to Y$ be a morphism of algebraic stacks.

If the geometric fibres of $f$ are algebraic spaces, then $f$ is representable by algebraic spaces.

I'm wondering about analogues of this fiberwise criterion in the context of gerbes and DM stacks.

For instance:

Q1. If the geometric fibres of $f$ are DM stacks, then is $f$ representable by DM stacks?

Q2. If there is a finite (abstract) group $G$ such that the geometric fibres of $f$ are $G$-gerbes, then is $f$ a $G$-gerbe?

• The answer to Q1 should be positive, as checking whether a morphism of algebraic spaces is (formally) unramified can be done on geometric points. – Ariyan Javanpeykar Jan 25 '16 at 22:21

As for Q2, I don't think so. For example, consider $(BG \times (\mathbb{A}^1\setminus \{ 0 \}))\amalg BG\to \mathbb{A}^1$, everything over $\mathbb{C}$.
Added [Edit: this one doesn't work, see comments below]: for a flat, but perhaps more contrived, example, I think one can take $\mathscr{X}:=B\mu_{3,\mathbb{Q}} \amalg B(\mathbb{Z}/3)_{\mathbb{Q}}$ and $X:=\mathrm{Spec}(\mathbb{Q})\amalg\mathrm{Spec}(\mathbb{Q})$, where $\mathscr{X}\to X$ is the coarse moduli space map. Here $\mu_{3,\mathbb{Q}}$ is the nonconstant group scheme over $\mathbb{Q}$ defined as a covariant functor on $\mathbb{Q}$-algebras by $A\mapsto$ third roots of $1$ in $A$, and $(\mathbb{Z}/3)_{\mathbb{Q}}$ is the constant group scheme defined on $\mathbb{Q}$-algebras by $A\mapsto\mathbb{Z}/3$ (Edit: this is for connected $Spec(A)$, in general it's a locally constant group scheme, as by the way it has to be a locally constant sheaf of groups). Now, I think the two basechanges along the two possible maps $\mathrm{Spec}(k)\to X$, $k$ algebraically closed, are both the gerbe $B\mu_{3,k}=B(\mathbb{Z}/3)_k$.
• This seems to work. The fibres are $BG$, right? The morphism is also smooth, no? – Marci Senf Jan 25 '16 at 18:38
• @Qfwfq You are right that $\mu_3$ are $\mathbb Z/3$ are non-isomorphic over $\mathbb Q$. Indeed, the group $\mu_3$ has no non-trivial sections over $\mathbb Q$. On the other hand, it is quite common for two neutral gerbes to be isomorphic even when the relevant group schemes are not. For a geometric example: consider the stack of smooth conics $B(PGL_2)$ over $\mathbb Q$. Let $X$ be a non-split smooth conic over $\mathbb Q$ with automorphism group $G$. Then $B(PGL_2)$ is isomorphic to $BG$, but $G$ is not isomorphic to $PGL_2$. (Here I write $PGL_2$ for what I should write $PGL_{2,\mathbb Q}$. – Ariyan Javanpeykar Jan 25 '16 at 22:20
• @Ariyan: interesting - can it even happen over $\mathbb{C}$? I mean, that $BG\cong BH$ for non isomorphic complex algebraic groups $G$ and $H$. – Qfwfq Jan 26 '16 at 12:45
• @Qfwfq That can certainly happen. Take $S$ to be the spectrum of a field (containing $\mathbb C$) such that there is a smooth conic $C$ over $S$ with no section. Let $G$ be the automorphism group scheme of $C$ over $S$. Then it follows that $BG = BPGL_{2,S}$, with $G$ not isomorphic to $PGL_{2,S}$. If you also want $S$ to be a finite type, then you will have to find a variety $S$ and a smooth conic with no section over $S$. This can be a bit more difficult (because of Tsen's theorem), but it should still be possible if $S$ is of dimension bigger than 1. – Ariyan Javanpeykar Jan 26 '16 at 16:42
• Just for fun: Example of a conic over $\mathbb{C}(s,t)$ with no rational point: $sx^2 + ty^2 = z^2$. – Daniel Loughran Jan 26 '16 at 17:55