I am reading https://arxiv.org/abs/0806.4160 to understand orbifolds as stacks.

Definition : Let $D\rightarrow Man$ be a stack over category of manifolds. An atlas for $D$ is a manifold $X$ and a map $p:X\rightarrow D$ such that for any map $f:M\rightarrow D$ from a manifold the fiber product $M\times_D X$ is a manifold and the map $pr_1:M\times_D X\rightarrow M$ is a surjective submersion.

I fail to understand what this means. I know what is a stack but I am not sure what it means to say a map from a manifold to a stack.

Any clarity would be welcome.

Before giving this definition, he says,

To keep the notation from getting out of control we drop the distinction between a manifold and the associated stack $\underline{M}$. We will also drop the distinction between stacks isomorphic to manifolds and manifolds.

This only made the definition complicated and not easier (for me) :D

Help me to understand the notion of atlas.

If it helps, I am trying to read about geometric stacks which are defined to be stacks over manifolds which possesses an atlas.

  • $\begingroup$ A morphism from a manifold $M$ to a stack $D$ is a functor from the slice $\underline M:=Man/M$ to $D$ over $Man$, i. e. such that the composite $Man/M\to D\to Man$ is equal to the projection $Man/M\to Man$. Well, more precisely not equal but isomorphic, and the choice of an isomorphism is part of the structure. $\endgroup$ – მამუკა ჯიბლაძე Jun 28 '18 at 7:38
  • $\begingroup$ @მამუკაჯიბლაძე Given a smooth manifold $M$, they have an associated groupoid $\underline{M}=\{M\rightrightarrows M\}$... Given any groupoid $\mathcal{G}$, the functor $B\mathcal{G}\rightarrow Man$ is a stack.. in particular, $B\underline{M}$ is a stack.. is this what you mean when you say slice? $\endgroup$ – Praphulla Koushik Jun 28 '18 at 8:03
  • $\begingroup$ Yes, except that not $B\underline M$ but the functor $B\underline M\to Man$ is a stack: according to 3.28 and 4.1, $B\underline M$ is equivalent to $Man/M$, the functor to $Man$ corresponding to the projection $Man/M\to Man$ ($(N\to M)\mapsto N$) $\endgroup$ – მამუკა ჯიბლაძე Jun 28 '18 at 8:43
  • $\begingroup$ Yes Yes, I mean the functor $B\underline{M}\rightarrow Man$ is a stack... I understand now @მამუკაჯიბლაძე :) $\endgroup$ – Praphulla Koushik Jun 28 '18 at 9:30
  • 1
    $\begingroup$ The idea is that a stack is locally a quotient of a manifold by some relation. This relation is exactly the Lie grupoid presenting the stack. This is a generalization of the disjoint product of charts covering a manifold. In this case, the Lie grupoid is the Cech grupoid. $\endgroup$ – user40276 Jul 6 '18 at 6:12

It is good to have a simple example in mind. An orbifold is a topological space for which each point $x$ has a neighborhood homeomorphic to the quotient of an open subset of $U_x$ of $\mathbb{R}^n$ by a finite group $G_x$. Here $X$ is $\coprod_xU_x$ and $p:X\rightarrow D$ is the restriction of the quotient map to $U_x$.

To be more concrete, if $D$ is the quotient of a finite group $G$ which acts on the manifold $X$ (with eventually fixed points) an atlas of $D$ is just $X$. To obtain an atlas of the orbifold you just have to blow-up the singularities.

You can also read the first section of my paper Differentiable Categories, gerbes and G-structures, Int. Journal of Contemp. Math. Sciences, Vol. 4, 2009, no. 29-32, 1547-1590, arXiv:0806.1357.

  • $\begingroup$ How does this answer my question? $\endgroup$ – Praphulla Koushik Jun 28 '18 at 9:16
  • $\begingroup$ Verify that the example in the answer satisfies the properties of your definition. $\endgroup$ – Tsemo Aristide Jun 28 '18 at 9:21
  • $\begingroup$ I was having trouble understanding the notations and definition.. Any ways, thank you for your post :) $\endgroup$ – Praphulla Koushik Jun 28 '18 at 9:32
  • $\begingroup$ You can have another approach to the subject in the first section of this paper: arxiv.org/pdf/0806.1357.pdf $\endgroup$ – Tsemo Aristide Jun 29 '18 at 12:21

An atlas for a stack $\mathcal{D}\rightarrow Man$ is

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

such that, given

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

the fibered product stack $\underline{M}\times_{\mathcal{D}}\underline{X}\rightarrow Man$ is isomorphic to a stack coming from a manifold and something extra happens.

As we are saying $\underline{M}\times_{\mathcal{D}}\underline{X}\rightarrow Man$ is coming from a manifold, we better give it a name. We denote that manifold by $M\times_{\mathcal{D}}X$.

This has nothing to do with fibered product of manifolds $M$ and $X$, of course we can not talk of fibered product of manifolds with respect to a category $\mathcal{D}$, it is just a notation that the stack is $\underline{M}\times_\mathcal{D}\underline{X}$ is isomorphic to the stack $\underline{M\times_\mathcal{D}X}$ coming from the manifold $M\times_\mathcal{D}X$.

Now, let us come to something extra happens.

This fibered prodcut $\underline{M}\times_\mathcal{D}\underline{X}$ comes with projection map $pr_1:\underline{M}\times_\mathcal{D}\underline{X}\rightarrow \underline{M}$.

enter image description here

Corollary $4.16$ in that article says :

Let $M,M'$ be two manifolds. For any map $F:\underline{M}\rightarrow \underline{M'}$ of categories fibered in groupoids there is a unique map of manifolds $f:M\rightarrow M'$ defining $f$.

So, for $pr_1:\underline{M}\times_\mathcal{D}\underline{X}=\underline{M\times_DX}\rightarrow \underline{M}$, there is a unique map associated $f:M\times_D X\rightarrow M$ and by something extra happens we mean to say this map $f:M\times_D X\rightarrow M$ is a submersion.

  • $\begingroup$ Yes, $f$ should be a surjective submersion. But I think that, to understand the heuristics, the submersion thing is a bit of a red herring. The point is that you want $f$ to be a cover in the Grothendieck topology you chose to put on $\mathbf{Man}$. It happens that surjective submersions give a Grothendieck topology on Man, but an equally good (maybe even equivalent, considered that in the smooth category submersions have sections...) Grothendieck topology is simply given by families of jointly surjective open embeddings. (...) $\endgroup$ – Qfwfq Jul 6 '18 at 16:54
  • $\begingroup$ (...) Now, such a jointly surjective family of open embeddings $\{j_i : U_i\to M\}$ is almost by definition an atlas for the manifold $M$. The "union" map $f:X:=\amalg_i U_i\to M$ is no longer an open embedding but it's still open, surjective, and a local diffeo; and it's essentially the same as giving the covering family $\{j_i\}$ (if you admit non connected manifolds with a lot of components in Man, then $\{f\}$ is a honest (singleton) cover for a still equivalent Grothendieck topology, given by -if I'm not mistaken- surjective local diffeos). $\endgroup$ – Qfwfq Jul 6 '18 at 16:55
  • $\begingroup$ @Qfwfq I think I understand something but not very sure.. you are saying condition on $f$ will vary depending on what Grothendieck topology I chose to put on Man.. I don’t think that article mentions anything about Grothendieck topology on Man, may be as they don’t want to discuss stack over arbitrary categories they did not care about grothendieck topology on Man.. $\endgroup$ – Praphulla Koushik Jul 6 '18 at 17:19
  • $\begingroup$ Yeah, let's say we have a site $(C,J)$, and a stack $Y$ over it. Then we may call $Y$ "geometric" if it has an atlas $f:U\to Y$, i.e. a $1$-morphism of stacks where $U$ is in $C$ and "locally on $Y$" $f$ is in $J$. By the last condition I mean that for every morphism $V\to Y$ where $V$ is in $C$, $U\times_{Y} V$ is in $C$ and the pullback map $U\times_{Y} V \to V$ is a $J$-cover. $\endgroup$ – Qfwfq Jul 6 '18 at 17:30
  • $\begingroup$ @Qfwfq ok ok. thanks. Can you give some reference where little more about geometric stacks is mentioned.. something al9ng the lines of what you said.. it looks very reasonable.. $\endgroup$ – Praphulla Koushik Jul 6 '18 at 17:35

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