# Stack being represented by a scheme/manifold

On page $$10$$ of the survey article Algebraic stacks, by T. Gomez (arXiv:math/9911199), we have following result

If a stack has an object with an automorphism other than the identity, then the stack can not be represented by a scheme.

Is the converse true?

Suppose I know only automorphism of any object are trivial automorphisms, then can I say that that stack is coming from a manifold/scheme.

Is there any way to construct such scheme/manifold.

Is it practical to check if each object has only trivial automorphisms.

Edit : There is some slight confusion regarding the notion of "no non trivial automorphisms". I want to clear that (mostly to myself).

A stack over a category(site) $$\mathcal{C}$$ can be seen in two ways.

One way as a category $$\mathcal{D}$$ with a functor $$\mathcal{D}\rightarrow \mathcal{C}$$ that is fibered in groupoids with some extra condition. Another way as a functor(almost) $$F:\mathcal{C}^{op}\rightarrow (\text{Gpd})$$ where $$\text{Gpd}$$ means the (2)category of groupoids.

To understand the notion of "no non trivial automorphisms" it is better (I think so) to see stack as $$F:\mathcal{C}^{op}\rightarrow (\text{Gpd})$$. By no non trivial automorphism for $$F$$ we mean the following :

Let $$U$$ be an object in $$\mathcal{C}$$. Then $$F(U)$$ is a groupoid. Let $$x,y\in X(U)$$ then, the set of arrows from $$x,y$$ denoted by $$\text{Hom}(x,y)$$ iseither empty or singleton. If this holds for each object $$U$$ of $$\mathcal{C}$$ and for each pair of objects $$x,y\in F(U)$$. Then, we say that this stack has no objects with non trivial automorphisms.

See a stack as a functor from category $$\mathcal{D}$$ to $$\mathcal{C}$$ say $$F:\mathcal{D}\rightarrow \mathcal{C}$$. Let $$U$$ be an object in $$\mathcal{C}$$. Look at the fiber category $$\mathcal{D}(U)$$. Let $$x,y$$ be objects in $$\mathcal{D}(U)$$ and $$Hom(x,y)$$ denote the arrows from $$x,y$$. If $$Hom(x,y)$$ is either singleton or empty for each pair of objects $$x,y$$ in $$\mathcal{D}(U)$$ and for each object $$U$$ in $$\mathcal{C}$$ then, we say the stack $$\mathcal{D}\rightarrow \mathcal{C}$$ has no non trivial automorphisms.

• This is definitely not true in the context of the article you quote --- essentially you are asking if all sheaves on the category of schemes (in your favorite subcanonical topology) are representable. There are many counterexamples; for example, there are algebraic spaces (which have an etale cover by representables) which are not schemes. Dec 2, 2018 at 19:19
• @DanielLitt I am familiar with differential geometric version of stacks and only recently started reading algebraic stacks. I do not know much about algebraic spaces... Even if you think this question is very trivial, I shamelessly request you to write (when you are free) little more details as as answer (only if you think this question is not off topic here)... Dec 2, 2018 at 19:26

• If all objects of a stack $S^{op} \to Gpd$ have trivial automorphism groups it is naturally isomorphic to a functor $S^{op} \to Set \hookrightarrow Gpd$, and since the stack satisfies descent, the functor $S^{op} \to Set$ satisfies descent and hence is a sheaf. No reference required. Dec 2, 2018 at 21:33
• Ok... there is some difference in notations.. Both of us are using base category as $S$... I am seeing stack as a category $\mathcal{D}$ with a functor $\mathcal{D}\rightarrow S$ with (among other) a condition that given an object $U$ of $S$ the fiber $\mathcal{D}(U)$ is a groupoid... This gives a functor $S^{op}\rightarrow Gpd$... This is the notation you are using... Ok.. Giving a Grothendieck topology on $S$ you can ask if this functor is a sheaf or not.. Dec 2, 2018 at 22:00
• @Praphulla Koushik: When people say "no nontrivial automorphisms" they mean $X(s)$ is equivalent to a rigid groupoid (one for which Hom(x,y) is either empty or a singleton) for every $s$. Dec 2, 2018 at 22:53
• @PK: consider the quotient groupoid of $\mathbb{R}^2$ by the vertical translation action of $\mathbb{R}$, $(x,y)\mapsto(x,y+v)$, $v\in\mathbb{R}$. The quotient -whatever it is- should be representable and equivalent to $\mathbb{R}$, so have no nontrivial automorphisms. Look at the object $s:=\{*\}$ (one point) of Manifolds. The groupoid $X(s)$ has a lot of objects: all the points of $\mathbb{R}^2$, but points $p=(x,y)$ and $p'=(x',y')$ on the same orbit are joined by the unique arrow corresponding to the unique $v\in\mathbb{R}$ for which $y'=y+v$. Points in distinct orbits aren't joined. Dec 2, 2018 at 23:11