# To check if a stack is coming from a manifold

Let $$\mathcal{D}$$ be a stack. An atlas for stack $$\mathcal{D}$$ is given by

• a smooth manifold $$X$$ and
• a map of stacks $$p:\underline{X}\rightarrow \mathcal{D}$$

such that, for any

• manifold $$M$$ and
• a map of stacks $$f:\underline{M}\rightarrow \mathcal{D}$$

the fiber product $$\underline{X}\times_{\mathcal{D}}\underline{M}$$ is represented by a manifold $$P$$ (i.e., it is isomorphic to stack $$\underline{P}$$) and the map $$P\rightarrow X$$ coming from the projection map $$\underline{X}\times_{\mathcal{D}}\underline{M}\rightarrow \underline{M}$$ is a smooth submersion.

I understand the set up.

What I do not understand is

1. How do you check if a map of stacks $$p:\underline{X}\rightarrow \mathcal{D}$$ is an atlas for $$\mathcal{D}$$?
2. How do you check if a stack $$\mathcal{C}$$ is isomorphic to $$\underline{P}$$ for some manifold.

Any comments are welcome.

Take $M = X$ and $f=p$, so that $\underline{P} = \underline{X} \times_\mathcal{D} \underline{X}$. The Lie groupoid $P\rightrightarrows X$ you get should be proper, in the sense the source-target map $(s,t)\colon P\to X\times X$ is a proper map, and $(s,t)$ should be injective, so that there are no nontrivial automorphism groups. Then your original stack is the stack associated to the manifold that is the quotient of $X$ by the equivalence relation corresponding to $P$ (which is closed, by properness of $(s,t)$), i.e. $\mathcal{D} \simeq \underline{X/P}$.
To respond to the question about non-injectivity of $(s,t)$ in the comments:
Let us say that the groupoid $P \rightrightarrows X$ has some pair $x,y\in X$ are such that $(s,t)^{-1}(x,y)$ has at least two distinct elements, say $a$ and $b$. Then note that $id_x$ and $b^{-1}a$ are distinct elements in $(s,t)^{-1}(x,x)=\mathrm{Aut}(x)$. Then consider the groupoid of functors $\ast \to (P\rightrightarrows X)$ and natural isomorphisms, which is equivalent to the groupoid of functors $\ast \to \mathcal{D}$ (this is because every principal $(P \rightrightarrows X)$-bundle over $\ast$ has a section - this is a nice exercise to work out the details). But by Yoneda the groupoid of functors $\ast \to \mathcal{D}$ is equivalent to the groupoid $\mathcal{D}(\ast)$, thinking of $\mathcal{D}$ as a functor $\mathbf{Mfld}^\mathrm{op}\to \mathbf{Gpd}$. But in the groupoid of functors $\ast\to (P\rightrightarrows X)$ the functor sending $\ast$ to $x\in X$ has a nontrivial automorphism, given by $b^{-1}a$, so the groupoid $\mathcal{D}(\ast)$ has an object with a nontrivial automorphism, hence $\mathcal{D}\colon \mathbf{Mfld}^\mathrm{op}\to \mathbf{Gpd}$ doesn't factor through $\mathbf{Set}\hookrightarrow \mathbf{Gpd}$. Every stack arising from a manifold is in fact a sheaf $\mathbf{Mfld}^\mathrm{op}\to \mathbf{Set}$, so in this case $\mathcal{D}$ cannot be the stack associated to a manifold. Hence for $\mathcal{D}$ to come from a manifold, $(s,t)$ must be injective. This is not sufficient, one really does need properness else the quotient $X/P$, which should be the manifold in question, will not exist in $\mathbf{Mfld}$. Consider for instance the action groupoid associated to a free action of $\mathbb{R}$ on a torus $U(1)^2$ with irrational slope.
• Thanks for your answer. I am trying to understand this.. Question is, given a geometric stack $\mathcal{D}$, how do you know if that stack comes from a manifold.. What you are saying is, take an atlas for $\mathcal{D}$ say $X\rightarrow \mathcal{D}$ and consider fiber product with itself, giving a stack, which we know is coming from a manifold, you are naming it $P$. You get a Lie groupoid $P\rightrightarrows X$.. – Praphulla Koushik Aug 15 '18 at 10:01
• If the map $(s,t)$ is proper and injective, then you are saying $\mathcal{D}$ comes from manifold $X/P$ where the relation is $x\sim y$ if there exists $p\in P$ such that $(s,t)(p)=(x,y)$.. Am I correct?? How did you think of this? Can you give some reference.. please.. – Praphulla Koushik Aug 15 '18 at 10:01