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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.

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    $\begingroup$ 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. $\endgroup$
    – Daniel Litt
    Commented Dec 2, 2018 at 19:19
  • $\begingroup$ @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)... $\endgroup$
    – Praphulla Koushik
    Commented Dec 2, 2018 at 19:26

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If all objects of a stack have trivial automorphism groups then it is equivalent to a sheaf, as pointed out by Daniel Litt in the comments. Pick your favourite non-representable sheaf as a counterexample to the claim that the stack is representable. For instance, on the site of manifolds, defined to be Hausdorff, paracompact, locally Euclidean topological spaces, consider the stack presented by a groupoid associated to an equivalence relation with non-Hausdorff quotient. In fact any equivalence relation such that the quotient is not a manifold will do (cf the irrational flow on a torus).

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  • $\begingroup$ Ok, if you can, please give reference for the result “If all objects of a stack have trivial automorphism groups then it is equivalent to a sheaf”... Assuming this, I think I got the rough idea.. there are non representable sheaves... so, that stack associated to sheaf which is not representable is not representable.. Ok.. $\endgroup$
    – Praphulla Koushik
    Commented Dec 2, 2018 at 21:18
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    $\begingroup$ 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. $\endgroup$
    – David Roberts ♦
    Commented Dec 2, 2018 at 21:33
  • $\begingroup$ 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.. $\endgroup$
    – Praphulla Koushik
    Commented Dec 2, 2018 at 22:00
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    $\begingroup$ @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 $s$. $\endgroup$
    – Qfwfq
    Commented Dec 2, 2018 at 22:53
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    $\begingroup$ @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. $\endgroup$
    – Qfwfq
    Commented Dec 2, 2018 at 23:11

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