Timeline for To check if a stack is coming from a manifold
Current License: CC BY-SA 4.0
34 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 17, 2019 at 7:10 | comment | added | Praphulla Koushik | Thank you for confirming. :) reference is not needed now.. :) I can prove for myself... | |
| Feb 17, 2019 at 6:56 | comment | added | David Roberts♦ | Is it this same Yes. | |
| Feb 17, 2019 at 6:23 | comment | added | Praphulla Koushik | The quotient map $X\rightarrow X/P$ is a surjective submersion.. We can pull back the groupoid $(X/P \rightrightarrows X/P)$ along this surjective submersion $X\rightarrow X/P$ (page $6$ of Non abelian differentiable gerbes).. So, we have groupoid $Q\rightrightarrows X$.Is it this same as $\mathcal{G}$? | |
| Feb 17, 2019 at 6:23 | comment | added | Praphulla Koushik | Can you give a reference for the proof of the result that $\mathcal{G}\cong (X/P\rightrightarrows X/P)$? Given conditions mentioned above we see that $X/P$ is a manifold... Is the equivalence $\mathcal{G}\cong (X/P\rightrightarrows X/P)$ some random equivalence or is the equivalence coming from pullback of groupoid $X/P\rightrightarrows X/P$ along surjective submersion $X\rightarrow X/P$?? | |
| Feb 17, 2019 at 6:02 | comment | added | Praphulla Koushik | Yes yes.. we don’t have $M$ before.. I did not mean I know $M$ before.. we only know that $P\rightrightarrows X$ is a proper lie groupoid and that $(s,t):P\rightarrow X\times X$ is injective.. it turns out that $X/P$ is a manifold and we have an equivalence $\mathcal{G}\rightarrow (X/P\rightrightarrows X/P)$ what I am trying to understand is how do we know these two are sufficient conditions to give equivalence | |
| Feb 16, 2019 at 22:00 | comment | added | David Roberts♦ | You don't fix $M$ in advance. It turns out that $M=X/P$. | |
| Feb 16, 2019 at 19:23 | comment | added | Praphulla Koushik | For Lie groupoid $M\rightrightarrows M$, $(s,t):M\rightarrow M\times M$ is just the diagonal map.. That is injective.. So, you want for $P\rightrightarrows X$, $(s,t):P\rightarrow X\times X$ to be injective map... So, it is very reasonable to ask for these two conditions namely $(s,t):P\rightarrow X\times X$ is proper and that $(s,t):P\rightarrow X\times X$ is injective... But, how do we knew that these are sufficient conditions to give a weak equivalence $(\mathcal{G}=P\rightrightarrows X)\rightarrow (M\rightrightarrows M)$?? | |
| Feb 16, 2019 at 19:18 | comment | added | Praphulla Koushik | One property is $M\rightrightarrows M$ is a proper Lie groupoid i.e., diagonal map $M\rightarrow M\times M$ is a proper map... So, you want $P\rightrightarrows X$ to be a proper Lie groupoid.... Isn't it so? I think this is your motivation for asking $P\rightrightarrows X$ to be a proper Lie groupoid... We thus have a proper map $(s,t):P\rightarrow X\times X$... This map gives an equivalence relation on $X$... As mentioned above, the relation is $x\sim y$ if there exists $p\in P$ such that $(s,t)(p)=(x,y)$... This relation is such that quotient space $X/P$ has a manifold structure... | |
| Feb 16, 2019 at 19:12 | comment | added | Praphulla Koushik | It looks more clearer now... I have a stack $\mathcal{D}$ and I want to see when this is representable by a manifold... This stack $\mathcal{D}$ gives a Lie groupoid $\mathcal{G}=P\rightrightarrows X$... You want to see an isomorphism of stacks $B\mathcal{G}\cong \underline{M}=B(M\rightrightarrows M)$... This happens only when these two Lie groupoids $\mathcal{G}$ and $(M\rightrightarrows M)$ are Morita equivalent (weak equivalent)... naturally you want $P\rightrightarrows X$ to have same properties that of $M\rightrightarrows M$... | |
| Sep 19, 2018 at 5:53 | comment | added | David Roberts♦ | $\ast$ is the category with one object and no non-identity arrows. | |
| Sep 19, 2018 at 5:27 | comment | added | Praphulla Koushik | I am not very sure if I understand your category $*$ correctly.. What is $*$ for you when you say consider groupoid of functors $*\rightarrow (P\rightrightarrows X)$ | |
| Sep 19, 2018 at 0:25 | history | edited | David Roberts♦ | CC BY-SA 4.0 |
Expanded answer
|
| Sep 18, 2018 at 22:49 | comment | added | Praphulla Koushik | If I write more here.. people say comments are not for extended discussion.. Can we continue this in a chatroom? | |
| Sep 18, 2018 at 22:39 | comment | added | David Roberts♦ | If (s,t) is not injective then there is an object of the groupoid with a non-trivial automorphism, which is the very hallmark of a non-trivial stack, and the entire motivation for the theory. It also means that that groupoid gives rise to a stack where one of the fibres is a groupoid not equivalent to a set, hence the stack cannot come from a manifold. | |
| Sep 18, 2018 at 17:02 | comment | added | Praphulla Koushik | I accepted this because I think this is very reasonable... | |
| Sep 18, 2018 at 17:01 | vote | accept | Praphulla Koushik | ||
| Sep 18, 2018 at 17:00 | comment | added | Praphulla Koushik | If $(s,t):P\rightarrow X\times X$ is as you have mentioned then, $X/P=X/\sim$ satisfies the condition in the link you have said math.stackexchange.com/questions/496571/… in answer by user43208 Thus, $X/P$ is a smooth manifold.. What is not clear is "if (s,t) is not injective you know instantly that you don't have a manifold as then you have genuinely stacky points" I do not understand this.. what do you mean by stack points.. I know it need not be a manifold in this case but could not understand what are stack points | |
| Sep 18, 2018 at 16:08 | comment | added | Praphulla Koushik | Suppose this Lie groupoid is such that the map $(s,t):P\rightarrow X\times X$ is proper and injective then $X/P$ is a manifold and there is an isomorphism of stacks $\mathcal{D}\rightarrow \underline{X/P}$... I am sorry, I was busy with some thing in last 20 days so could not spend time on this.. | |
| Sep 18, 2018 at 16:07 | comment | added | Praphulla Koushik | I am just saying to get more clarity... I have a geometric stack $\mathcal{D}$ and I want to prove that it comes from a manifold... I have to choose an atlas for $\mathcal{D}$ say $p:\underline{X}\rightarrow \mathcal{D}$... As this is an atlas, $\underline{X}\times_{\mathcal{D}}\underline{X}$ is a manifold.. Say $P$... Consider the Lie groupoid $P\rightrightarrows X$ that comes from obvious projection maps $\underline{X}\times_{\mathcal{D}}\underline{X}\rightrightarrows \underline{X}$... | |
| Aug 22, 2018 at 2:38 | comment | added | Praphulla Koushik | It looks very surprising that for any manifold $M$ and a submersion $U\rightarrow M$, though the Lie groupoids $\{M\rightrightarrows M\}$ and $\{U\times_MU\rightrightarrows U\}$ are not equivalent, their associated stacks $B(M\rightrightarrows M)$ and $B(U\times_M U\rightrightarrows U)$ are equivalent... I mean it is not necessary that if say $B\mathcal{G}$ and $B\mathcal{H}$ are equivalent then $\mathcal{G}$ and $\mathcal{H}$ are equivalent, we can only have $\mathcal{G}$ and $\mathcal{H}$ to be Morita equivalent.. This is some way of saying about Localization I guess.. | |
| Aug 21, 2018 at 23:10 | comment | added | David Roberts♦ | @PraphullaKoushik "By every manifold gives a Lie groupoid you mean the Lie groupoid..." <-- yes. "Are these two notions..." <-- they give the same stack, up to equivalence. This is the whole point of the localisation business, going back to Pronk's work in the 1990s. These Lie groupoids are not equivalent in the usual sense, which is why we want to formally invert some functors if we take Lie groupoids as the objects under consideration. If we only work with the associated stacks, then they are already equivalent. | |
| Aug 21, 2018 at 19:38 | comment | added | Praphulla Koushik | @DavidRoberts Suppose there is a surjecive submersion $U\rightarrow M$ such that the Cech groupoid (submersion groupoid) $\{U\times_M U\rightrightarrows U\}$ is categorically equivalent to $X$ then do we say $X$ is coming from a manifold? Are these two notions, $\{M\rightrightarrows M\}$ and $\{U\times_M U\rightrightarrows U\}$ same?? I could not see why they are same.. | |
| Aug 21, 2018 at 19:36 | comment | added | Praphulla Koushik | @DavidRoberts I do not know how I missed your comment. I thought you did not respond.. By every manifold gives a Lie groupoid you mean the Lie groupoid $\{M\rightrightarrows M\}$ for a manifold $M$, right?? I can ask what it means for a Lie groupoid $X$ to be equivalent to manifold $M$ i.e., I can ask when there is an equivalence of categories $\{M\rightrightarrows M\}\rightarrow X$.. I am confused why notion of Cech groupoid is coming here... | |
| Aug 16, 2018 at 22:18 | comment | added | David Roberts♦ | Every manifold gives a Lie groupoid, and ao you can ask what it means for a Lie groupoid X to be equivalent to a manifold M in the localised bicategory, equivalently, what it means to have a functor C(U)--> X that is fully faithful and essentially surjective, where C(U) is the Cech groupoid corresponding to a surjective submersion U-->M | |
| Aug 16, 2018 at 17:08 | comment | added | Praphulla Koushik | I am trying to prove on my own the result that you have mentioned... I could not see what Lie groupoid equivalence to manifold is you are referring in that 2012 paper... I have deleted that question $\underline{M}\rightarrow B\mathcal{G}$ being given by a principal $\mathcal{G}$ bundle over $M$ on some one's advice... | |
| Aug 15, 2018 at 17:36 | comment | added | Praphulla Koushik | @DavidRoberts thanks for reference. I will see that.. :) | |
| Aug 15, 2018 at 17:03 | comment | added | David Roberts♦ | @Praphulla Sorry, I thought that would have been clear from the content. It's this one tac.mta.ca/tac/volumes/26/29/26-29abs.html (and I forgot the other paper was officially published in 2012!) | |
| Aug 15, 2018 at 13:53 | comment | added | Willie Wong | @DavidRoberts: as an aside: do you mind retagging this question with some appropriate top-order (arxiv) tags? Currently it is only tagged [[stacks]]. | |
| Aug 15, 2018 at 12:02 | comment | added | Praphulla Koushik | Oh, ok ok.. Can you tell me title of your 2012 paper.. there seem to be two... | |
| Aug 15, 2018 at 10:28 | comment | added | David Roberts♦ | Also, if (s,t) is not injective you know instantly that you don't have a manifold as then you have genuinely stacky points. | |
| Aug 15, 2018 at 10:17 | comment | added | David Roberts♦ | The geometric stack is equivalent to a manifold if the Lie groupoid I gave is equivalent to a manifold in the sense in my 2012 paper. (This is because the 2-categories of geometric stacks and internal groupoids with anafunctors as maps are equivalent.) That is the case precisely when (s,t) is injective and the quotient of the resulting equivalence relation is a manifold. This is the case precisely when (s,t) is closed (then properness comes for free), a result going back to Godement, I believe (see eg math.stackexchange.com/q/496571/3835 ) | |
| Aug 15, 2018 at 10:01 | comment | added | Praphulla Koushik | 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.. | |
| Aug 15, 2018 at 10:01 | comment | added | Praphulla Koushik | 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 $P\rightrightarrows X$.. | |
| Aug 15, 2018 at 6:06 | history | answered | David Roberts♦ | CC BY-SA 4.0 |