# 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. – Daniel Litt Dec 2 '18 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)... – Praphulla Koushik Dec 2 '18 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. – David Roberts Dec 2 '18 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.. – Praphulla Koushik Dec 2 '18 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 $s$. – Qfwfq Dec 2 '18 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. – Qfwfq Dec 2 '18 at 23:11