Timeline for Fibered product of stacks comes from a Lie groupoid
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39 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 15, 2019 at 11:52 | vote | accept | Praphulla Koushik | ||
| Feb 15, 2019 at 11:51 | comment | added | Praphulla Koushik | Oh.. :) I was banging my head to wall to understand why it is relevant... This is my last comment for today as I promised... I will ask some question tomorrow... :) Please leave a message when you are free (anytime today or tomorrow) for five minutes... | |
| Feb 15, 2019 at 11:00 | comment | added | David Roberts♦ | It's irrelevant, it was just a comment. | |
| Feb 15, 2019 at 6:35 | comment | added | Praphulla Koushik | Assuming $\theta$ is surjective, I want to see what is the isotropy group. I see that isotropy group $(\mathcal{G}\times_{\mathcal{H}}\mathcal{G})_h=G\times_H G$. What I do not understand is, why do we bother about this $G\times_H G$ being Lie subgroup of $G\times G$?? | |
| Feb 15, 2019 at 6:35 | comment | added | Praphulla Koushik | I want to see when this Lie groupoid is transitive Lie groupoid so that I can then say this Lie groupoid is weakly equivalent to a Lie groupoid of the form $(K\rightrightarrows *)$ for some Lie group $K$, more precisely this $K$ is the isotropy group of some object of $(\mathcal{G}\times_{\mathcal{H}}\mathcal{G})$. I then see that $\theta$ is surjective then Lie groupoid is transitive and if it is transitive then $\theta$ is surjective. | |
| Feb 15, 2019 at 6:33 | comment | added | Praphulla Koushik | What I am trying to understand here is... given a morphism of Lie groups $G\rightarrow H$, which gives a morphism of Lie groupoids $F:(G\rightrightarrows *)\rightarrow (H\rightrightarrows *)$, what is the $2$-fibre product of $F$ with itself i.e., what is $\mathcal{G}\times_{\mathcal{H}}\mathcal{G}$?? I understand that the groupoid $\mathcal{G}\times_{\mathcal{H}}\mathcal{G}$ is $G\times H\times G\rightrightarrows H$ where source map is $s(g_1,h,g_2)=h\cdot\theta(g_1)$ and target map is $t(g_1,h,g_2)=\theta(g_2)\cdot h$. | |
| Feb 15, 2019 at 6:32 | comment | added | Praphulla Koushik | In short, my question is the following : Why do we bother about this $G\times_H G$ being Lie subgroup of $G\times G$?? There is already a Lie group structure on $G\times_H G$, just like any Isotropy group of Lie groupoid... (mathoverflow.net/questions/301638/…).. If the question is not sensible, I am explaining below more carefully... | |
| Feb 15, 2019 at 6:14 | comment | added | Praphulla Koushik | :) :) No No... I do not expect you to reply immediately.... I will just leave comment here.. Please see only when you are super free.... That chat thing I pressed only because it is saying again and again to move to chat... I will write here only :) :) No problem at all... Please carry on with your work and see when you are free :) | |
| Feb 15, 2019 at 6:13 | comment | added | David Roberts♦ | @PraphullaKoushik sorry, but not now. | |
| Feb 15, 2019 at 2:39 | comment | added | David Roberts♦ | I don't know what you're asking. | |
| Feb 14, 2019 at 22:51 | comment | added | Praphulla Koushik | As isotrpy group is a Lie group, this (after identification) is a Lie subgroup of $G\times H\times G$ which does not say if $G\times_H G$ is Lie subgroup of $G\times G$... but, why do we need that? I do not know what I am missing... All you want is a Lie group and $G\times_H G$ is a Lie group (not necessarily Lie subgroup of $G\times G$).. It will be good if it is a Lie subgroup of $G\times G$ but why do we even need that? I do not know what I am misunderstanding.. | |
| Feb 14, 2019 at 22:47 | comment | added | Praphulla Koushik | One small (last) question... Even though maps are different, I got same result. $\theta: G\rightarrow H$ is surjective iff Lie groupoid is transitive.Transitive Lie groupoid is weak equvalent to isotropy group..So, I should just pick some random $h\in H$ and see what is its $\mathcal{G}_h$... Here also I got same thing.. Istropy group is $G\times_HG$... It is ok till here... | |
| Feb 14, 2019 at 22:20 | comment | added | David Roberts♦ | @PraphullaKoushik I may have written down an isomorphic groupoid :-) | |
| Feb 14, 2019 at 21:55 | comment | added | Praphulla Koushik | Yes, I am doing that now... There seem to be slight difference in source/target maps of Lie groupoid $G\times H\times G\rightarrow H$ for what you said and what I got... I see $s:G\times H\times G\rightarrow H$ is given by $(g_1,h,g_2)\mapsto h\cdot \theta(g_1)$ and $t:G\times H\times G\rightarrow H$ is given by $(g_1,h,g_2)\mapsto \theta(g_2)\cdot h$... I do not think it makes so much difference... The next thing I will ask is when this Lie groupoid is transitive.. I think I might end up with same conlcusion as you... | |
| Feb 14, 2019 at 21:10 | comment | added | David Roberts♦ | @PraphullaKoushik no, it's not written in a book or paper that I know of, I just gave you an outline of a proof. You should fill in any details you feel are not obvious in your own private notes. | |
| Feb 14, 2019 at 14:52 | comment | added | Praphulla Koushik | Is this written some where? In a paper or a book or something like that? It is not that I did not understand and wanting to see somewhere... I understand this... I can write down in detail here as an answer but as always it will be like I am writing for my self and it does not make any difference for others.. :D :D My interest is, If this is written in a paper or something like that I am hoping I can see more results like this in that source... | |
| Feb 14, 2019 at 14:47 | comment | added | Praphulla Koushik | One small/quick question... Given morphisms of Lie groupoids $\phi:\mathcal{G}\rightarrow \mathcal{K}$ and $\psi:\mathcal{H}\rightarrow \mathcal{K}$ , under some condition, the $2$-fibre product $\mathcal{G}\times_{\mathcal{K}}\mathcal{H}$ is a Lie groupoid. In a particular case when $\mathcal{G}=\mathcal{H}=(G\rightrightarrows *)$ and $\mathcal{K}=(H\rightrightarrows *)$ for Lie groups $G,H$, you have said above a situation when the fibre product $\mathcal{G}\times_{\mathcal{K}}\mathcal{H}$ is also a Lie group of the form $(K\rightrightarrows *)$.. | |
| Feb 2, 2019 at 12:28 | comment | added | Praphulla Koushik | Definitely reasonable thing to say... :) :) | |
| Feb 2, 2019 at 10:46 | comment | added | David Roberts♦ | But any weak equivalence $\mathcal{H}\rightarrow (\mathcal{G}_x\rightrightarrows *)$ is an actual equivalence, since $H_0 \to \ast$ has a section. And in any case, we are supplied with the inclusion functor, which is more than one can say for the abstract general case. | |
| Feb 2, 2019 at 6:48 | comment | added | Praphulla Koushik | By too much, I mean the following... For stacks $B\mathcal{G}$ and $B\mathcal{G_x}$ to be isomorphic, we only need that $\mathcal{G}$ and $\mathcal{G}_x$ are Morita equivalent Lie groupoids (page $6$ of Moerdijk) i.e., it is sufficient if there exist some Lie geoupoid $\mathcal{H}$ with weak equivalence $\mathcal{H}\rightarrow \mathcal{G}$ and $\mathcal{H}\rightarrow (\mathcal{G}_x\rightrightarrows *)$.. Asking inlcusion functor to be weak equivalence seem to be more than sufficient here.. so, was mentioning that... | |
| Feb 2, 2019 at 6:23 | comment | added | Praphulla Koushik | I forgot to mention.. As you said, there is no obvious map $(X_1\rightrightarrows X_0)\rightarrow (X_x\rightrightarrows *)$ but there is this inclusion map $(X_x\rightrightarrows *)\rightarrow (X_1\rightrightarrows X_0)$ given by $g:x\rightarrow x$ mapping to $g$ and $*$ is mapping to $x\in X_0$... And you want this functor to be a weak equivalence.. If this holds then we are done but I somehow think this is too much to ask.. I will figure out what is going on here... | |
| Feb 2, 2019 at 6:12 | comment | added | Praphulla Koushik | Thansk for clarification :) I am ok with weak equivalence :) I directly jumped into section 8 and was confused.. I will read from beginning of the paper (1 page per day is my plan) and see if I can grasp something.... I also decided not to ask you more than one clarifications/suggestion/help per day... | |
| Feb 2, 2019 at 5:46 | comment | added | David Roberts♦ | Right? <--- yes. The stuff in section 8 of my paper relies on reading some definitions in previous sections, to see why Lie groupoids are an example :-) "Morita equivalence" here is a phrase with multiple definitions in this context, so I find it an unhelpful phrase. Better is "weak equivalence", since that refers to an actual functor. You don't want a weak equivalence $(X_1\rightrightarrows X_0)\rightarrow (X_x\rightrightarrows *)$, since such a thing might not exist, but the inclusion functor $(X_x\rightrightarrows *) (X_1\rightrightarrows X_0)\rightarrow $ to be a weak equivalence. | |
| Feb 2, 2019 at 3:43 | comment | added | Praphulla Koushik | So, here we want $t\pi_1:\mathcal{G}_1\times_{\mathcal{G}_0}\mathcal{H}_0\rightarrow \mathcal{G}_0$ i.e., $t\circ \pi_1:\mathcal{G}_1\times \mathcal{G}_0\rightarrow \mathcal{G}_0$ to be surjective submersion... (Orbifolds as groupoids, Moerdijk, page 5, subsection 2.4) This is what you are saying... Right? Do you want me to see anything else in that paper? I could not understand what you are referring to in your paper.. CAn you please confirm.. | |
| Feb 2, 2019 at 3:31 | comment | added | Praphulla Koushik | Are you saying for $B\mathcal{G}=B\mathcal{G}_x$, we need a Morita equivalence of categories $(\mathcal{G}_1\rightrightarrows \mathcal{G}_0)\rightarrow (\mathcal{G}_x\rightrightarrows *)$... A morphism of Lie groupoids $\mathcal{G}\rightarrow \mathcal{H}$ is a Morita equivalence if 1) $t\pi_1:\mathcal{G}_1\times_{\mathcal{G}_0}\mathcal{H}_0\rightarrow \mathcal{G}_0$ is a surjective submersion 2) some other condition.... | |
| Feb 2, 2019 at 3:24 | comment | added | Praphulla Koushik | I just wanted to confirm that I understand your last comment completely.... I said, for transitive Lie groupoids $\mathcal{G}$, we have that $\mathcal{G}_x$ is a Lie group and that $B\mathcal{G}=B\mathcal{G}_x$... Then you said "you also need $t:s^{-1}(x)\rightarrow \mathcal{G}_0$ is a submersion" and asked me to look at your paper, section 8... I could not figure out what you are referring in that paper.. | |
| Jan 21, 2019 at 6:08 | comment | added | David Roberts♦ | I am trying to find a proof of that result. <-- it's the definition of a weak equivalence $X\to Y$ of Lie groupoids. See my paper tac.mta.ca/tac/volumes/26/29/26-29abs.html especially the discussion in section 8 | |
| Jan 21, 2019 at 2:37 | comment | added | Praphulla Koushik | I am sorry if “you changed that question” sounded rude... I was thinking it was your strategy... Yes, you described $K$ in this particular case, I was thinking if there is any common technique to get such Lie group... I understand that $K$ you have explained.. “you also need $t:s^{-1}(x)\rightarrow \mathcal{G}_0$ to be a submersion”... ok ok.. I am trying to find a proof of that result. Thank you. | |
| Jan 21, 2019 at 1:13 | history | edited | David Roberts♦ | CC BY-SA 4.0 |
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| Jan 21, 2019 at 1:12 | comment | added | David Roberts♦ | You have changed this question to "Given a morphism of Lie groupoids ... <--- no, I only addressed the case when you have Lie groups $G$ and $H$ and a map of stacks $BG\to BH$. Every such map is isomorphic to one coming from a homomorphism $G\to H$. || So, you are saying If this particular groupoid ...is a transitive Lie groupoid <-- Yes. || Is there any way how to get this $K$ <-- I described it in my answer. || I am guessing for a transitive Lie groupoid.. <-- almost: you also need $t\big|_{s^{-1}(x)} \colon s^{-1}(x) \to Obj(\mathcal{G})$ to be a submersion. | |
| Jan 20, 2019 at 15:22 | comment | added | Praphulla Koushik | I know (mathoverflow.net/questions/301638/…) that Isotropy group $\mathcal{G}_x$ of any object $x\in \mathcal{G}_0$ is a Lie group.. So, I am guessing for a transitive Lie groupoid $\mathcal{G}$ we have $B\mathcal{G}=B\mathcal{G}_x$ i.e., $\mathcal{G}$ is Morita equivalent to a Lie groupoid coming from Lie group $\mathcal{G}_x$... This is just my guess that $\mathcal{G}$ is Morita equivalent to $\mathcal{G}_x$... Is this correct? Is this what you are thinking when you said above thing or am I misunderstanding something? | |
| Jan 20, 2019 at 15:19 | comment | added | Praphulla Koushik | You said "For the stack $BG\times_{BH} BG$ to be equivalent to one of the form $BK$ for some topological group $K$, the topological groupoid I just described must be transitive"... This I am not very sure how to take this.. So, you are saying If this particular groupoid $G\times H\times G\rightrightarrows H$ is a transitive Lie groupoid, then, there exists a Lie group $K$ such that $B(G\times H\times G\rightrightarrows H)\cong BK$... Is there any way how to get this $K$? | |
| Jan 20, 2019 at 15:04 | comment | added | Praphulla Koushik | So, the question was "Given a morphism of stacks $B\mathcal{G}\rightarrow B\mathcal{H}$, when does the fibre product $B\mathcal{G}\times_{B\mathcal{H}}B\mathcal{G}$ a stack coming from Lie groupoid"... You have changed this question to "Given a morphism of Lie groupoids $\mathcal{G}\rightarrow \mathcal{H}$, when does the fibred product $\mathcal{G}\times_{\mathcal{H}}\mathcal{G}$ is a Lie groupoid"... Then, you are saying if $\mathcal{G}\times_{\mathcal{H}}\mathcal{G}$ is a Lie groupoid, then $$B\mathcal{G}\times_{B\mathcal{H}}B\mathcal{G}\cong B(\mathcal{G}\times_{\mathcal{H}}\mathcal{G})$$. | |
| Jan 20, 2019 at 12:47 | comment | added | Praphulla Koushik | Here $\mathcal{H}\times_{\mathcal{G}}\mathcal{K}$ given that you have morphism of Lie groupoids $\mathcal{H}\rightarrow \mathcal{G}$ and $\mathcal{H}\rightarrow \mathcal{K}$.... To be precise both morphisms of Lie groupoids for you is same thing and that is $\{G\rightrightarrows *\}\rightarrow \{H\rightrightarrows *\}$... Ok.. Thank you :) There is also an explanation in arxiv.org/pdf/math/0203100.pdf says when $\mathcal{H}\times_{\mathcal{G}}\mathcal{K}$ is a Lie groupoid.. That looks same thing as what you said. Thanks for that thesis Link... I am sure that is useful... | |
| Jan 20, 2019 at 12:42 | comment | added | Praphulla Koushik | :) I was thinking this is similar to pull back of a Lie groupoid $\mathcal{G}_1\rightrightarrows \mathcal{G}_0$ along a smooth map $J:M\rightarrow \mathcal{G}_0$ which can be found in page no 6 section $2.2$ in arxiv.org/pdf/math/0511696.pdf... There is also a notion of fibre product of $\mathcal{H}\times_{\mathcal{G}}\mathcal{K}$ given two morphism of Lie groupoids $\mathcal{H}\rightarrow \mathcal{G}$ and $\mathcal{K}\rightarrow \mathcal{G}$ which can be found in page $5$ section $2.3$ named Fibered products in arxiv.org/pdf/math/0203100.pdf | |
| Jan 20, 2019 at 12:23 | comment | added | David Roberts♦ | It is the weak pullback of $(G\rightrightarrows \ast) \to (H \rightrightarrows \ast)$ along itself. See Definition 1.15 in arxiv.org/pdf/1512.04209.pdf | |
| Jan 20, 2019 at 11:53 | comment | added | Praphulla Koushik | I am still trying to guesss how you thought of the Lie groupoid $G\times H\times G\rightrightarrows H$... Can you please tell me how did this occur suddenly? | |
| Jan 20, 2019 at 10:31 | comment | added | Praphulla Koushik | About Lie groups, yes we need $G\times_H G$ to be a Lie subgroup of $G\times G$.. You can see edit version 1 mathoverflow.net/revisions/320210/1 I assumed it is a Lie subgroup... Thanks. I will respond after spending some time on this. :) | |
| Jan 20, 2019 at 10:16 | history | answered | David Roberts♦ | CC BY-SA 4.0 |