# Fibered product of stacks comes from a Lie groupoid

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 $$K$$?

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.

• 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 10:31
• 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 11:53
• 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 12:23
• :) 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 12:42
• 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 10:46

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.)

• 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 $BK$... Jan 7, 2019 at 6:25
• @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. Jan 7, 2019 at 16:42
• 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 $\mathcal{K}$.. Jan 7, 2019 at 17:16
• @PraphullaKoushik: Yes, in the example that I gave the pullback is not a Lie groupoid. Jan 7, 2019 at 17:19
• 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. Jan 7, 2019 at 17:22