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I am adding some context here. I am reading Introduction to Differentiable Stacks by Gregory Ginot.

In page no $7$, just before the remark $2.2$ he says the following.

One shall be careful that fibre product of differentiable stacks is not necessarily a differentiable stack (though it is a stack over $\text{Diff}$).

So, one question we can think of is,

when does a fibre product of differentiable stacks a differentiable stack?

This seems to be too difficult to answer. So, I look for a simpler question.

When does $B\mathcal{G}\times_{B\mathcal{H}} B\mathcal{G}$ a differentiable stack where the fibre product is coming from a given morphism of stacks $B\mathcal{G}\rightarrow B\mathcal{H}$.


Question in context of Lie groups is as follows.

Suppose $G,H$ be Lie groups and $F:BG\rightarrow BH$ be a map of stacks.

When can we say that the fibered product $BG\times_{BH}BG$ is a differentiable stack i.e., of the form $B\mathcal{K}$ for some Lie groupoid $\mathcal{K}$?

Assume that this $F:BG\rightarrow BH$ is coming from a morphism of Lie groups $\theta:G\rightarrow H$.

Then, we can ask following question.

When can we say that the fibered product $BG\times_{BH}BG$ is a differentiable stack coming from Lie group i.e., of the form $BK$ for some Lie group $K$?

Any comments are welcome.

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Pullbacks of stacks coming from Lie groupoids are not always equivalent to Lie groupoids.

Take $G=H=\mathbb{R}$. Define $F(x)=0$ if $x\leq 0$ and $F(x)=exp(−1/x^2)$ if $x>0$.

The pullback is not equivalent to a Lie groupoid in this situation: the set-theoretical pullback is $(−\infty,0]\times(−\infty,0]\cup \{(x,x)|x\in \mathbb{R}\}$, which is clearly not a smooth manifold.

One can guarantee that the pullback is a Lie groupoid by imposing transversality conditions on the maps involved. That is, $A \times_C B$ is a Lie groupoid if $A_0 \rightarrow C_0 \leftarrow B_0$ is transversal and $A_1 \rightarrow C_1 \leftarrow B_1$ is transversal, where subscripts $0$ and $1$ denotes objects and morphisms respectively. (In fact, this transversality condition guarantees that the pullback is also a homotopy pullback, which is almost always what one actually wants.)

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  • $\begingroup$ 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 $BK$... $\endgroup$ – Praphulla Koushik Jan 7 at 6:25
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    $\begingroup$ @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. $\endgroup$ – Dmitri Pavlov Jan 7 at 16:42
  • $\begingroup$ 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 $\mathcal{K}$.. $\endgroup$ – Praphulla Koushik Jan 7 at 17:16
  • $\begingroup$ @PraphullaKoushik: Yes, in the example that I gave the pullback is not a Lie groupoid. $\endgroup$ – Dmitri Pavlov Jan 7 at 17:19
  • $\begingroup$ 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. $\endgroup$ – Praphulla Koushik Jan 7 at 17:22
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Think of $BG$ and $BH$ as topological stacks, whereby one can calculate a topological groupoid presenting the stack $BG\times_{BH} BG$, namely the following: the object space is the space underlying $H$ and the morphism space is $G\times H \times G$. The source map $s\colon G\times H \times G \to H$ is the projection on the middle factor; the target map $t\colon G\times H \times G \to H$ is $(g_1,h,g_2) \mapsto \theta(g_1)^{-1}h\,\theta(g_2)$. (The reason you can do this is because the Lie groupoids presenting the stacks $BG$ and $BH$ only have a single object, so the situation is rather special. For more general Lie groupoids it is a little more fiddly, but not too different.)

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: every object should be isomorphic to every other object. That is, for every pair $h_1,h_2\in H$ there should be elements $g_1,g_2\in G$ such that $h_2 = \theta(g_1)^{-1}h_1\theta(g_2)$. If $\theta$ is surjective, then you can take $g_1 = e_G$ and $g_2$ any lift of $h_1^{-1}h_2$, so this is a sufficient condition (in the topological case).

Now if we go back to general $\theta$, but take the special case $h_1 = e_H$, then we require for any $h\in H$ that $h= \theta(g_1^{-1}g_2)$, so in fact $\theta$ surjective is a necessary condition. To figure out which group $K$ it is such that $BG\times_{BH} BG \simeq BK$—it is a priori well-defined up to isomorphism of topological groups—we need to consider the pairs $g_1,g_2$ such that $h = \theta(g_1)^{-1}h\theta(g_2)$ for some chosen $h\in H$. We might as well take $h=e_H$, so that we want pairs $g_1,g_2 \in G$ such that $e_H = \theta(g_1^{-1}g_2)$, or in other words, such that $\theta(g_1) = \theta(g_2)$. The space of such pairs is just $G\times_H G$, which is a topological subgroup of $G\times G$, and so $K=G\times_H G$.

(In fact $G\times \ker \theta \to G\times_H G$, $(g,k) \mapsto (g,gk)$ is a homeomorphism, but only a topological group isomorphism if $\ker \theta \lt G$ is central subgroup.)


Now if we want to do this in Lie groups, then everything works, except that we need $K = G\times_H G$ to be a sub-Lie-group of $G\times G$, and this is so if $\theta$ is a submersion. But a surjective map of (finite-dimensional) Lie groups is automatically a submersion. Thus $G\to H$ is a surjective submersion, and hence a locally trivial bundle (this follows from using charts derived from the exponential map and the surjective map of the associated Lie algebras).

If we don't care about $BG\times_{BH} BG \simeq BK$ for some $K$, then $\theta$ being a submersion should be enough to make $G\times H \times G \rightrightarrows H$ a Lie groupoid (the only hard part is to show that $(g_1,h,g_2) \mapsto \theta(g_1)^{-1}h\,\theta(g_2)$ is a submersion).

Even in the special case analysed above, we don't know that $K$ is a central subgroup, or that $G\twoheadrightarrow H$ is a central extension, but that's not necessary for your question. So we find ourselves in the situation Dmitri gave in greater generality: $(G\rightrightarrows \ast) \to (H \rightrightarrows \ast)$ is a submersion on arrows and object components.


Added There was a small unimportant lie in what I wrote: it is not sufficient in the topological case for $\theta$ to be just surjective, to have an equivalence of stacks $BG \times_{BH} BG \simeq BK$. Namely, it is not true that the groupoid that I described being transitive is enough. One must have a certain map to have local sections, and this boils down to requiring that $\theta$ have local sections. In the smooth case we ask that this map is a surjective submersion, but this follows from $\theta$ just being surjective, as this already implies it is a submersion.

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  • $\begingroup$ 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. :) $\endgroup$ – Praphulla Koushik 2 days ago
  • $\begingroup$ 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? $\endgroup$ – Praphulla Koushik 2 days ago
  • $\begingroup$ 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 $\endgroup$ – David Roberts 2 days ago
  • $\begingroup$ :) 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 $\endgroup$ – Praphulla Koushik 2 days ago
  • $\begingroup$ 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... $\endgroup$ – Praphulla Koushik 2 days ago
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Question is

Given a morphism of stacks $B\mathcal{G}\rightarrow B\mathcal{H}$ when does the fibre product $B\mathcal{G}\times_{B\mathcal{H}}B\mathcal{G}$ is of the form $B\mathcal{K}$ for some Lie groupoid $\mathcal{K}$?

I did not mention that this morphism of stacks $B\mathcal{G}\rightarrow B\mathcal{H}$ comes from a morphism of Lie groupoids $\mathcal{G}\rightarrow \mathcal{H}$ but Dmitri Pavlov assumed that this the case and that is very reasonable assumption. I should have mentioned that this map of stacks $B\mathcal{G}\rightarrow B\mathcal{H}$ comes from a morphism of Lie groupoids $\mathcal{G}\rightarrow \mathcal{H}$.

Given a morphism of Lie groupoids $\mathcal{G}\rightarrow \mathcal{K}$ and a morphism of Lie groupoids $\mathcal{H}\rightarrow \mathcal{K}$ there is a notion of fibre product $\mathcal{G}\times_{\mathcal{K}}\mathcal{H}$ in page no $5$ in Orbifolds as Groupoids: an Introduction by Ieke Moerdijk. It is not necessary for $\mathcal{G}\times_{\mathcal{K}}\mathcal{H}$ to be a Lie groupoid always. There is a criterion that says when $\mathcal{G}\times_{\mathcal{K}}\mathcal{H}$ is a Lie groupoid. Once we know that it is Lie groupoid, it is expected that $$B\mathcal{G}\times_{B\mathcal{K}}B\mathcal{H}\cong B(\mathcal{G}\times_{\mathcal{K}}\mathcal{H})$$ I could not prove this, I think this is true.

So, fibre product of map of stacks $B\mathcal{G}\times_{B\mathcal{K}}B\mathcal{H}$ comes from a Lie groupoid under some conditions. The condition i sthat $\mathcal{G}_0\times_{\mathcal{K}_0}\times\mathcal{K}_1\times_{\mathcal{K}_0}\mathcal{H}_0$ is a manifold.

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