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Given a Lie group $G$, we have a Lie groupoid $(G\rightrightarrows *)$ and stack $BG=B\mathcal{G}$ of principal $G$ bundles.

Given a smooth manifold $M$, we have Lie groupoid $(M\rightrightarrows M)$ and stack $\underline{M}$ whose objects are smooth maps to $M$.

Given Lie group $G$, we have two Lie groupoids associated to it :

  • $(G\rightrightarrows *)$ if we consider Lie group structure.
  • $(G\rightrightarrows G)$ if we ignore group structure and treat it as a manifold.

We have corresponding stacks associated :

  • $(G\rightrightarrows *)$ gives stack $B(G\rightrightarrows *)$, usually denoted by $BG$.
  • $(G\rightrightarrows G)$ gives stack $B(G\rightrightarrows G)$, usually denoted by $\underline{G}$.

As any Lie group is a manifold, shouldn't there be some relation with notions $BG$ and $\underline{G}$? I see they are not same. How are they related?

It is not even the case that the Lie groupoid $(G\rightrightarrows *)$ is pull back of $(G\rightrightarrows G)$ or the otherway around.


I do not know counter part in Algebraic geometry.

Feel free to (I request you to) relate this to algebraic geometry version of stacks.

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    $\begingroup$ Can some one up voting the question leave a message :P :D $\endgroup$ Commented Feb 20, 2019 at 18:03
  • $\begingroup$ I don't understand what is the utility of "Are they same? I see they are not same." $\endgroup$
    – LSpice
    Commented Feb 20, 2019 at 18:13
  • $\begingroup$ @LSpice I have edited it. :) Bad English skills :) $\endgroup$ Commented Feb 20, 2019 at 18:17
  • $\begingroup$ It's nothing to do with algebraic geometry, just pure stack theory. $\endgroup$
    – David Roberts
    Commented Feb 20, 2019 at 21:29
  • $\begingroup$ @DavidRoberts I though similar question can be asked from algebraic geometry perspective so that it will be convenient for them to think.. $\endgroup$ Commented Feb 21, 2019 at 4:52

1 Answer 1

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$\underline{G}$ is the homotopy loop space of $BG$.

More precisely, the two terminal maps $G\rightarrow pt$ and $G\rightarrow pt$ yield a weak equivalence $\underline{G} \rightarrow pt\times_{BG} pt$, where the right side denotes the homotopy pullback of the diagram $pt\rightarrow BG\leftarrow pt$ and $pt$ denotes the representable stack of a smooth manifold given by a single point.

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  • $\begingroup$ I will say what I understand.. For Lie group $G$, we have obvious map of stacks $\underline{*}\rightarrow BG$ (which is actually an atlas for $BG$). Consider the $2$-fibre product $\underline{*}\times_{BG}\underline{*}$. Here, $*$ denote singleton space and $\underline{*}$ denote the stack associated to singleton manfold. What does it mean to say a weak equivalence $G\rightarrow \underline{*}\times_{BG}\underline{*}$.. I only know weak (Morita ??) equivalence in case of Lie groupoids.. Does it mean weak equivalence as in page 9 of maths.qmul.ac.uk/~noohi/papers/quick.pdf? $\endgroup$ Commented Feb 20, 2019 at 19:34
  • $\begingroup$ searching in google said that, "loop space" of space $BG$ is a space that is homotopy equivalent to $G$ (math.stackexchange.com/questions/442805/…) Can you please tell me How should I relate with my question? $\endgroup$ Commented Feb 20, 2019 at 19:47
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    $\begingroup$ Where by "homotopy pullback" Dmitri means the pullback of stacks in the appropriate 2-categorical way (or, if you like, the comma object). The loop space here is not the same as the topological loop space (though there is a relation there too), but a higher-categorical analogue (cf ncatlab.org/nlab/show/loop+space+object). "Weak equivalence" means Morita equivalence of the corresponding Lie groupoids, but as I've said elsewhere, I don't think "Morita equivalence" is the best phrase to use. $\endgroup$
    – David Roberts
    Commented Feb 20, 2019 at 21:25
  • $\begingroup$ @PraphullaKoushik: What David Roberts said. In addition, be aware that BG in the hyperlinked question is not BG in your question: the former BG is a space (alias ∞-groupoid), whereas your BG is a Lie groupoid, or, alternatively, a stack of ∞-groupoids on the site of smooth manifolds. The former BG is the shape of the latter, see ncatlab.org/nlab/show/shape+modality. $\endgroup$ Commented Feb 20, 2019 at 23:07
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    $\begingroup$ Yes, that's what his answer says. $\endgroup$
    – David Roberts
    Commented Feb 21, 2019 at 5:42

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