Skip to main content
29 events
when toggle format what by license comment
Feb 21, 2019 at 11:14 vote accept Praphulla Koushik
Feb 21, 2019 at 9:43 answer added David Roberts timeline score: 1
Feb 21, 2019 at 8:39 comment added Praphulla Koushik Ok :) @DavidRoberts :)
Feb 21, 2019 at 7:15 comment added David Roberts Perhaps I should have written $S^1$ instead of $U(1)$. It's meant to just be a map stacks representable by manifolds.
Feb 21, 2019 at 6:30 comment added Praphulla Koushik @DavidRoberts I do not know why I am misunderstanding like this... thanks.. I will spend some more time...
Feb 20, 2019 at 21:21 comment added David Roberts Then the diagonal map (by Yoneda) is precisely that arising from the ordinary map of manifolds (so not as a Lie group).
Feb 20, 2019 at 17:36 comment added Praphulla Koushik It is true that given manifolds $M,N$ if $\underline{M}\rightarrow \underline{N}$ is an epimorphism then $M\rightarrow N$ is a surjective submersion... I do not think it holds for the case of Lie groups... I have proved, given any morphism of Lie groups $G\rightarrow H$, associated map $BG\rightarrow BH$ is an epimorphism... I can show you if you permit me to show.. @DavidRoberts
Feb 19, 2019 at 20:28 comment added David Roberts Epimorphism of manifolds = surjective function
Feb 19, 2019 at 16:58 comment added Praphulla Koushik I am trying to understand your last comment for surjectivity.. You are saying that, if $G\rightarrow H$ is not surjective then, $BG\rightarrow BG\times_{BH}BG$ need not be an epimorphism You said to consider $\left<e \right>\rightarrow U(1)$ and consider $\left<e\right>\rightarrow \left<e\right>\times _{BU(1)}\left<e\right>$.. This is same as the $\left<e\right>\rightarrow U(1)$.. you said this is not surjective. That is ok, but we do not need it to be surjective.. we want it to be an epimorphism.. Am I misunderstanding something.we need to look for epimor., not surjective...@DavidRoberts
Jan 21, 2019 at 6:14 comment added Praphulla Koushik @DavidRoberts Ok, Ok. I will work on that...Thank you.
Jan 21, 2019 at 6:08 comment added David Roberts See my answer at mathoverflow.net/a/321296/4177, I don't know any simpler proof off the top of my head.
Jan 21, 2019 at 2:43 comment added Praphulla Koushik Yes yes, definitely.. I wanted to say for surjective submersion... @DavidRoberts is there a quick proof for the fact that $B\hat{G}\times_{BG}B\hat{G}\cong B(\hat{G}\times_G\hat{G})$ given that I know $\hat{G}\rightarrow G$ is a surjective submersion..
Jan 20, 2019 at 22:32 comment added David Roberts "It is true that..." this uses the fact that $\hat{G}\to G$ is surjective. || "$BG\to BH$ is a gerbe..." well, in the previous comment you used surjectivity, so it can't be for arbitrary homomorphisms. Take the case that $H=U(1)$ and $G=\langle e \rangle = pt$ and see if $pt \to pt\times_{BU(1)} pt$ is an epimorphism of stacks: it is not, since $pt\times_{BU(1)} pt$ is just the stack associated to the underlying manifold of $U(1)$. Then the diagonal map (by Yoneda) is precisely that arising from the ordinary map of manifolds, $e\colon \pt \to U(1)$—not surjective!
Jan 20, 2019 at 18:55 comment added Praphulla Koushik @DavidRoberts As $BG$ is trivially a stack (gerbe for some people), the obvious map of stacks $BG\rightarrow BH$ coming from a morphism of Lie groups should be a gerbe over stack... Does this sentence make some sense? I am having a feeling that $BG\rightarrow BH$ is a gerbe over stack for any morphism of Lie groups $G\rightarrow H$... SAme thing that you have said...
Jan 20, 2019 at 13:47 comment added Praphulla Koushik Now, the diagonal map $B\hat{G}\rightarrow B\hat{G}\times_{BG}B\hat{G}$ would then be simply arising from $\hat{G}\rightarrow \hat{G}\times_G\hat{G}$.. and I have seen that for any morpshim of Lie groups $G\rightarrow H$, $BG\rightarrow BH$ is an epimorphism.. So, for $\hat{G}\rightarrow \hat{G}\times_G\hat{G}$ the corresponding map $B\hat{G}\rightarrow B(\hat{G}\times_G\hat{G})=B\hat{G}\times_{BG}B\hat{G}$ is an epimorphsim.. Thus, $B\hat{G}\rightarrow BG$ is a gerbe over stack. I am not able to see the flaw. @DavidRoberts
Jan 20, 2019 at 13:44 comment added Praphulla Koushik It is true that $B\hat{G}\times_{BG}B\hat{G}=B(\hat{G}\times_G\hat{G})$ (as $\hat{G}\rightarrow G$ is a submersion, the pull back $\hat{G}\times_G\hat{G}$ is an embedded submanifold of $\hat{G}\times \hat{G}$.. It is also true that $\hat{G}\times_G\hat{G}$ is a subgroup (closed under multiplication and inverse). Any subgroup that is an embedded submanifold is a Lie subgroup. So, $\hat{G}\times_G\hat{G}$ is a Lie subgroup of $\hat{G}\times\hat{G}$)... @DavidRoberts
Jan 19, 2019 at 21:01 comment added David Roberts have you checked that $[pt/\hat{G}] \to [pt/\hat{G}] \times_{[pt/G]} [pt/\hat{G}]$ is an epimorphism?
Jan 19, 2019 at 19:04 comment added Praphulla Koushik @DavidRoberts I think what you said is true... I am able to see that given a morphism of Lie groups $\theta:G\rightarrow H$, the corresponding map of stacks $[*/G]\rightarrow [*/H]$ is a gerbe over stack (epimorphism for sure, I checked it just now and I do not need the condition that it is a principal bundle. For principal bundle it is true but it is not necessary)... This seems very surprising.. Why did Heinloth mentioned that there is some central extension $S^1\rightarrow \hat{G}\rightarrow G$??
Jan 14, 2019 at 3:49 comment added Praphulla Koushik I woke up just now (10 am for me... I worked till 5 am, thanks to you :D.. you said there is a hope, so I worked harder.. though i did not get the result in the generality you said but I got something)... I will write down after some time if you want to see @DavidRoberts
Jan 14, 2019 at 3:38 comment added Praphulla Koushik @DavidRoberts that’s true.. just the bundle $U\times \hat{G}\rightarrow U$.. I tried previously also the same thing but left at a point where I thought there is no hope.. yesterday night I tried and it worked to some extent.. I did not wanted to say anything in hurry.. so written down in detail.. it seems like I need to assume $\hat{G}\rightarrow G$ is a principal $S^1$ bundle.. if you want to see what I have done, I can show.. How much you are sure that it has nothing to do with $\hat{G}\rightarrow G$ is a principal $S^1$ bundle? It does not seem to work for any morphism of Lie groups..
Jan 14, 2019 at 3:05 comment added David Roberts Any trivial $G$-bundle $U\times G$ has a canonical lift to a principal $\hat{G}$-bundle :-)
Jan 13, 2019 at 20:43 comment added Praphulla Koushik @DavidRoberts Thanks, I am trying.. You said "just take a trivialising cover for $P$." I tried with same technique before but I could not succeed.. I will try one more time.... :)
Jan 13, 2019 at 20:33 comment added David Roberts yes ........ :-)
Jan 13, 2019 at 20:30 comment added Praphulla Koushik @DavidRoberts Ok. Thanks for the clarification :) First I have to prove that $[*/\hat{G}]\rightarrow [*/G]$ is a gerbe over stack and then prove that pull back (fiber product) of gerbe over stack is a gerbe over stack (it may be obvious but I did not prove yet).. It looks like it has nothing to do with central extension... Any morphism of Lie groups $\hat{G}\rightarrow G$ such that there is a local section $U\rightarrow \hat{G}$ should give a gerbe over stack $[*/\hat{G}]\rightarrow [*/G]$... Is that the case?
Jan 13, 2019 at 20:25 comment added David Roberts In response to your comment: yes.
Jan 13, 2019 at 20:24 comment added David Roberts Then, also, it should be easier to see why the stack $[pt/\hat{G}]$ of principal $\hat{G}$-bundles is a gerbe over $[pt/G]$. Given any $X\to [pt/G]$, that is, a principal $G$-bundle $P\to X$, there is a cover $U\to X$ such that $U\to X \to [pt/G]$ lifts to $[pt/\hat{G}]$: just take a trivialising cover for $P$. Thus $[pt/\pi]$ is an epimorphism of stacks. A similar type of thinking—unwinding the definition of the stack in terms of bundles—will help to show that $[pt/\hat{G}] \to [pt/\hat{G}] \times_{[pt/G]} [pt/\hat{G}]$ is also an epimorphism.
Jan 13, 2019 at 20:12 comment added Praphulla Koushik @DavidRoberts I am getting confused with English ... You are asking to take $X=pt$ and then saying consider $[X/G]\rightarrow [pt/G]$... Are you saying consider the obvious map of stacks $[X/G]\rightarrow [*/G]$ and pull back $[*/\hat{G}]\rightarrow [*/G]$ along $[X/G]\rightarrow [*/G]$ to get $[X/\hat{G}]\rightarrow [X/G]$?? As $[*/\hat{G}]\rightarrow [*/G]$ is a gerbe over stack, so is the pull back $[X/\hat{G}]\rightarrow [X/G]$?? Is this what you mean?
Jan 13, 2019 at 20:05 comment added David Roberts You can consider the special case of $X=pt$, since the example you give is pulled back along $[X/G] \to [pt/G]$, and the pullback of a gerbe is a gerbe.
Jan 13, 2019 at 18:43 history asked Praphulla Koushik CC BY-SA 4.0