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.

  • 1
    $\begingroup$ Can some one up voting the question leave a message :P :D $\endgroup$ 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
    Feb 20, 2019 at 18:13
  • $\begingroup$ @LSpice I have edited it. :) Bad English skills :) $\endgroup$ Feb 20, 2019 at 18:17
  • $\begingroup$ It's nothing to do with algebraic geometry, just pure stack theory. $\endgroup$
    – David Roberts
    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.. $\endgroup$ Feb 21, 2019 at 4:52

1 Answer 1


$\underline{G}$ 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.

  • $\begingroup$ 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? $\endgroup$ 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? $\endgroup$ Feb 20, 2019 at 19:47
  • 1
    $\begingroup$ 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
    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. $\endgroup$ Feb 20, 2019 at 23:07
  • 3
    $\begingroup$ Yes, that's what his answer says. $\endgroup$
    – David Roberts
    Feb 21, 2019 at 5:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.