Timeline for Fibered product of stacks comes from a Lie groupoid
Current License: CC BY-SA 4.0
31 events
| when toggle format | what | by | license | comment | |
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| Jan 30, 2019 at 14:36 | history | bounty ended | Praphulla Koushik | ||
| Jan 22, 2019 at 3:38 | comment | added | Praphulla Koushik | Thank you, I will read that. | |
| Jan 21, 2019 at 23:22 | comment | added | Dmitri Pavlov | @PraphullaKoushik: All pullbacks are limits. A good list of references for homotopy limits is available on the nLab: ncatlab.org/nlab/show/homotopy+limit. Preservation of limits by the Yoneda embedding is discussed here: ncatlab.org/nlab/show/… | |
| Jan 21, 2019 at 16:28 | comment | added | Praphulla Koushik | I got confused because you are saying Transversal limit.. By your above comment it is clear that you mean Transversal pullback which is defined in answer... please suggest some place to read about homotopy limits and the result that Yoneda embedding preserves homotopy limits... I am new to this so do not know why Yoneda embedding preserves homotopy limits I.e., why $B\mathcal{G}\times_{B\mathcal{K}}B\mathcal{H}=B(\mathcal{G}\times_{\mathcal{K}}\mathcal{H})$ please suggest a reference.. apologies for many questions.. could not find it on searching online.. | |
| Jan 21, 2019 at 16:08 | comment | added | Dmitri Pavlov | @PraphullaKoushik: Transversal pullbacks are defined in my answer. A pullback that is transversal is automatically also a homotopy pullback. The Yoneda embedding (denoted B in your posts) preserves homotopy limits, in particular, homotopy pullbacks. Thus, applying B to a transversal pullback square of Lie groupoids produces a homotopy pullback square of stacks, which is precisely what you asked for in you second-to-the-last comment above (BG×BKBH≅B(G×KH)). | |
| Jan 21, 2019 at 14:45 | comment | added | Praphulla Koushik | I am sorry, I am not yet comfortable with notion of Transversal limits of Lie groupoids and homotopy limits.. I am guessing by Transversal limit you mean the pull back groupoid of Lie groupoids intersecting Transversally.. Is that correct? I do not understand at all when you say "In particular, the image of a transversal pullback square of Lie groupoids is a homotopy pullback square of stacks in groupoids" Pardon me for my dumbness.. Please explain little more.. | |
| Jan 20, 2019 at 19:33 | comment | added | Dmitri Pavlov | @PraphullaKoushik: Transversal limits of Lie groupoids are also homotopy limits, and the Yoneda embedding preserves homotopy limits. In particular, the image of a transversal pullback square of Lie groupoids is a homotopy pullback square of stacks in groupoids. | |
| Jan 20, 2019 at 16:19 | comment | added | Praphulla Koushik | That is true, I was slightly confused... I realised it late... You are saying that, if I have morphism of Lie groupoids $\mathcal{G}\rightarrow \mathcal{K},\mathcal{H}\rightarrow \mathcal{K}$ intersect transversally as you said, then the fibre product $\mathcal{G}\times_{\mathcal{K}}\mathcal{H}$ is a Lie groupoid.. This is ok at the level of Lie groupoids.. Is there any way to see this at the level of stacks.. Is there a quick proof that in case Lie groupoids intersecting transversally then $B\mathcal{G}\times_{B\mathcal{K}}B\mathcal{H}\cong B(\mathcal{G}\times_{\mathcal{K}}\mathcal{H})$?? | |
| Jan 20, 2019 at 16:10 | comment | added | Dmitri Pavlov | @PraphullaKoushik: The case of Lie groups is a special case of the criterion given in the answer: the object maps are ←*→, which is transversal, so you only have to ask that the morphisms of Lie groups are transversal. | |
| S Jan 20, 2019 at 8:09 | history | suggested | Praphulla Koushik | CC BY-SA 4.0 |
latex edit to make it easily readable
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| Jan 20, 2019 at 4:53 | review | Suggested edits | |||
| S Jan 20, 2019 at 8:09 | |||||
| Jan 20, 2019 at 4:49 | comment | added | Praphulla Koushik | I have edited my question to make it clearer.. I unaccepted the answer as the case of Lie groups is still not clear... | |
| Jan 20, 2019 at 4:34 | vote | accept | Praphulla Koushik | ||
| Jan 20, 2019 at 4:47 | |||||
| Jan 7, 2019 at 22:28 | vote | accept | Praphulla Koushik | ||
| Jan 20, 2019 at 4:26 | |||||
| Jan 7, 2019 at 18:47 | comment | added | Dmitri Pavlov | @PraphullaKoushik: This is a legitimate question about pullbacks of Lie groupoids with a legitimate answer. Deleting it goes against the policies of this site. Even if it was deleted, it would be pretty soon undeleted because there is nothing wrong with it. | |
| Jan 7, 2019 at 18:04 | comment | added | Praphulla Koushik | I think I have to take a break.. i am one even able to see what is written there.. apologies.. can you please delete your answer so that I can delete this question.. | |
| Jan 7, 2019 at 18:01 | comment | added | Dmitri Pavlov | @PraphullaKoushik: The second paragraph in my answer explains carefully what transversality for Lie groupoids means: the object maps must be transversal and the morphism maps must be transversal (as maps of smooth manifolds). | |
| Jan 7, 2019 at 17:59 | comment | added | Praphulla Koushik | What is transversal intersection in arbitrary lie groupoids? I know what is it in case of manifolds maps but not in case of morphism of Lie groupoids...You still have that code even when some one edits it. Why not leave it like that.. | |
| Jan 7, 2019 at 17:56 | comment | added | Dmitri Pavlov | @PraphullaKoushik: I answered the new version with arbitrary Lie groupoids. I do not use MathJax because I need to retain the ability to copy-paste my texts. | |
| Jan 7, 2019 at 17:56 | comment | added | Praphulla Koushik | Not every Lie groupoid is of the form of manifold.. what can be said about General Lie groupoid. You are referring to Lie groupoids coming from a manifold,. | |
| Jan 7, 2019 at 17:54 | history | edited | Dmitri Pavlov | CC BY-SA 4.0 |
answered the new version of the question
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| Jan 7, 2019 at 17:39 | comment | added | Praphulla Koushik | Oh no, I do not know why I said they are Lie groups.. for me $G,H$ means Lie groups... ok ok.. I would not have got confused if it was $M$ and $N$.. :) You are treating smooth manifolds as Lie groupoids.. that is fine.. I think I did not ask the question clearly.. With your permission, I would like to delete this question and ask a better one...this as it is seems to be not useful for anyone.. | |
| Jan 7, 2019 at 17:35 | comment | added | Dmitri Pavlov | @PraphullaKoushik: There are no Lie groups in my answer. G and H are smooth manifolds, hence also Lie groupoids. I simply reused the notation from the first part of your question. If you need a positive result for some special case (which is not what your original question asks for), you can always demand transversality of the involved maps, which guarantees that the pullback is a Lie groupoid. | |
| Jan 7, 2019 at 17:22 | comment | added | Praphulla Koushik | You want me to treat Lie group $G$ as a Lie groupoid.. that’s fine.. I am looking for a positive result in some special case.. I am aware that pullback need not be manifold.. Is it clear what I am trying to ask? My English is not good so I don’t know if it conveyed correctly. | |
| Jan 7, 2019 at 17:19 | comment | added | Dmitri Pavlov | @PraphullaKoushik: Yes, in the example that I gave the pullback is not a Lie groupoid. | |
| Jan 7, 2019 at 17:18 | history | edited | Dmitri Pavlov | CC BY-SA 4.0 |
Clarified the nonexistence statement
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| Jan 7, 2019 at 17:16 | comment | added | Praphulla Koushik | So, you are saying for $F:B\mathcal{G}\rightarrow B\mathcal{H}$ be a morphism of stacks the fibered product $B\mathcal{G}\times_{B\mathcal{H}}B\mathcal{G}$ is not necessarily of the form $B\mathcal{K}$ for some Lie groupoid $\mathcal{K}$.. | |
| Jan 7, 2019 at 16:42 | comment | added | Dmitri Pavlov | @PraphullaKoushik: You have two sentences marked with a question mark. The "first claim" refers to the first of these two questions. So it is false that the pullback is a Lie groupoid. | |
| Jan 7, 2019 at 6:25 | comment | added | Praphulla Koushik | I did not understand what you mean by "Your first claim"... All you are saying is that pullback of a smooth map $G\rightarrow H$ is not a manifold... Are you saying something else also... I understand that given a smooth map $G\rightarrow H$, the pullback $G\times_HG$ is not necessarily a manifold... I am looking for a criterion on $G\rightarrow H$ and also on $BG\rightarrow BH$ so that $G\times_HG$ is a Lie group $K$ and $BG\times_{BH}BG$ is the space $BK$... | |
| Jan 6, 2019 at 21:01 | history | edited | Thomas Rot | CC BY-SA 4.0 |
added 46 characters in body
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| Jan 6, 2019 at 20:27 | history | answered | Dmitri Pavlov | CC BY-SA 4.0 |