Let $\mathcal{G}$ be a Lie groupoid. The target map $t:\mathcal{G}_1\rightarrow \mathcal{G}_0$ is a principal $\mathcal{G}$ bundle.

This article Orbifolds as Stacks? by Eugene Lerman calls (in page $11$) this particular principal bundle to be the unit principal $\mathcal{G}$ bundle. So, for a Lie groupoid $\mathcal{G}$, this $\mathcal{G}$ bundle is a special element in $B\mathcal{G}$.

Let $\mathcal{G}$ and $\mathcal{H}$ be Lie groupoids. For $B\mathcal{G}$, I have a special element $t:\mathcal{G}_1\rightarrow \mathcal{G}_0$. For $B\mathcal{H}$, I have a special element $t:\mathcal{H}_1\rightarrow \mathcal{H}_0$. We can ask the follwing question.

Are "the unit elemnts" preserved by an arbitrary map of stacks $F:B\mathcal{G}\rightarrow B\mathcal{H}$ i.e., do we always have $F(t:\mathcal{G}_1\rightarrow \mathcal{G}_0)=t:\mathcal{H}_1\rightarrow \mathcal{H}_0$ for any map of stacks $F:B\mathcal{G}\rightarrow B\mathcal{H}$.

This question does not make sense. As $F$ preserves fibers, $F(t:\mathcal{G}_1\rightarrow \mathcal{G}_0)$ would be a principal $\mathcal{H}$ bunle of the form $Q\rightarrow \mathcal{G}_0$. So, $F(t:\mathcal{G}_1\rightarrow \mathcal{G}_0)$ which is of the form $Q\rightarrow \mathcal{G}_0$ can not be the principal $\mathcal{H}$ bundle $t:\mathcal{H}_1\rightarrow \mathcal{H}_0$.

We have principal $\mathcal{H}$ bundles $Q\rightarrow \mathcal{G}_0$ and $\mathcal{H}_1\rightarrow \mathcal{H}_0$. As base spaces are different there is no way (in general) to compare these two principal $\mathcal{H}$ bundles.

Suppose we are in a situation where $B\mathcal{G}\rightarrow B\mathcal{H}$ is coming from a morphism of Lie groupoids $\mathcal{G}\rightarrow \mathcal{H}$. Realising that there is an obvious map between base spaces of these two bundles namely $\phi:\mathcal{G}_0\rightarrow \mathcal{H}_0$ makes the question of looking for some relation between $Q\rightarrow \mathcal{G}_0$ and $\mathcal{H}_1\rightarrow \mathcal{H}_0$ more specific. The question would then be,

Is $Q\rightarrow \mathcal{G}_0$ same the pull back of $t:\mathcal{H}_1\rightarrow \mathcal{H}_0$ along $\phi:\mathcal{G}_0\rightarrow \mathcal{H}_0$.

In a very special case when this morphism of Lie groupoids $\phi:\mathcal{G}\rightarrow \mathcal{H}$ is a Lie groupoid extension i.e., when $\phi:\mathcal{G}_0\rightarrow \mathcal{H}_0$ is an identity map (and some thing extra), we can ask

Is $Q\rightarrow \mathcal{G}_0$ same thing as $t:\mathcal{H}_1\rightarrow \mathcal{H}_0=\mathcal{G}_0$.

I could not see why this is true from definition of map of stacks but I feel this should be the case. Any comments are welcome.

Edit : A stack (over the category of manifolds $\text{Man}$) for me is a category $\mathcal{D}$ along with a functor $\mathcal{D}\rightarrow \text{Man}$ such that it is a category fibered in groupoids and some extra conditions. By an element of stack I mean an element (object) in the category $\mathcal{D}$.

Consider the special case when this map of stacks $B\mathcal{G}\rightarrow B\mathcal{H}$ is coming from a morphism of Lie groupoids $\mathcal{G}\rightarrow \mathcal{H}$. I will recall how one gets $B\mathcal{G}\rightarrow B\mathcal{H}$ from $\mathcal{G}\rightarrow \mathcal{H}$.

We first construct a $\mathcal{G}-\mathcal{H}$ bibundle given $\phi:\mathcal{G}\rightarrow \mathcal{H}$. We consider principal $\mathcal{H}$ bundle $t:\mathcal{H}_1\rightarrow \mathcal{H}_0$, pull it back along $\phi:\mathcal{G}_0\rightarrow \mathcal{H}_0$ to get a principal $\mathcal{H}$ bundle, now with the base $\mathcal{G}_0$ which is precisely $\mathcal{G}_0\times_{\mathcal{H}_0}\mathcal{H}_1\rightarrow \mathcal{H}_0$. The manifold $\mathcal{G}_0\times_{\mathcal{H}_0}\mathcal{H}_1$ has an action of $\mathcal{G}$ on it, given by $g.(x,h)=(t(g),\phi(g)h)$. With this action, it becomes a $\mathcal{G}-\mathcal{H}$ bibundle.

enter image description here

This $\mathcal{G}-\mathcal{H}$ bibundle gives morphism of stacks $B\mathcal{G}\rightarrow B\mathcal{H}$ the map is given by composition of bibundle. The object $\mathcal{G}_1\rightarrow \mathcal{G}_0$ is mapped to $\left(\mathcal{G}_1\times_{\mathcal{G}_0}(\mathcal{G}_0\times_{\mathcal{H}_0}\mathcal{H}_1)\right)/\mathcal{G}_1\rightarrow \mathcal{G}_0$ which comes from following diagram

enter image description here

We have $\left(\mathcal{G}_1\times_{\mathcal{G}_0}(\mathcal{G}_0\times_{\mathcal{H}_0}\mathcal{H}_1)\right)/\mathcal{G}_1= \mathcal{G}_0\times_{\mathcal{H}_0}\mathcal{H}_1$.

So, the principal bundle is $\mathcal{G}_0\times_{\mathcal{H}_0}\mathcal{H}_1\rightarrow \mathcal{G}_0$.

If $\mathcal{G}\rightarrow \mathcal{H}$ is a Lie groupoid extension, then we have $\mathcal{H}_0=\mathcal{G}_0$ and $\mathcal{G}_0\times_{\mathcal{H}_0}\mathcal{H}_1=\mathcal{H}_1$. So, $\mathcal{G}_0\times_{\mathcal{H}_0}\mathcal{H}_1\rightarrow \mathcal{G}_0$ is just the target map $t:\mathcal{H}_1\rightarrow \mathcal{H}_0$.

So, if $F:B\mathcal{G}\rightarrow B\mathcal{H}$ is given by a morphism of Lie groupoids $\mathcal{G}\rightarrow \mathcal{H}$ which is a Lie groupoid extension, then $F(t:\mathcal{G}_1\rightarrow M)=t:\mathcal{H}_1\rightarrow M$.

Otherwise, it does not make sense to ask if $F$ takes unit principal bundle of $\mathcal{G}$ to unit principal bundle of $\mathcal{H}$.

  • What do you mean by an element of a stack? For me a stack is a sheaf of groupoids, and the principal bundle $t:\mathcal G_1\to\mathcal G_0$ is an object in the groupoid $B\mathcal G(\mathcal G_0)$. A morphism of stacks $B\mathcal G\to\mathcal BH$ sends this to an object of $B\mathcal H(G_0)$, and if you present the morphism as a functor $F_i:\mathcal G_i\to\mathcal H_i$, you can check that this object is the principal $\mathcal H$-bundle $\mathcal H_1\times_{t,\mathcal H_0,F_0}\mathcal G_0$ (the map $F_i$ and functoriality is used to turn this into a principal bundle). – Bertram Arnold Dec 4 at 11:53
  • I think you mean to say "[---] If you present the morphism as a functor $F:\mathcal{G}\rightarrow \mathcal{H}$ [---]"... @BertramArnold – Praphulla Koushik Dec 4 at 12:42
  • There is something here which does not make sense. As any map of stacks $B\mathcal{G}\rightarrow B\mathcal{H}$ is fiber preserving, it takes an object $t:\mathcal{G}_1\rightarrow \mathcal{G}_0$ in $B\mathcal{G}$ to some thing of the form $Q\rightarrow \mathcal{G}_0$ which can not be $t:\mathcal{H}_1\rightarrow \mathcal{H}_0$. I was having in mind the map of stacks coming from a Lie groupid extension (which is simply a morphism of Lie groupoids with same base i.e., $\mathcal{G}_0=\mathcal{H}_0$ with some extra condition on $\phi:\mathcal{G}_1\rightarrow \mathcal{H}_1$). – Praphulla Koushik Dec 4 at 13:13
  • In that case, object $t:\mathcal{G}_1\rightarrow \mathcal{G}_0$ goes to some thing of the form $Q\rightarrow \mathcal{G}_0$ in $B\mathcal{H}$. As $\mathcal{H}_0=\mathcal{G}_0$, object $t:\mathcal{G}_1\rightarrow \mathcal{G}_0$ goes to some thing of the form $Q\rightarrow \mathcal{H}_0$ in $B\mathcal{H}$. Now, I can ask if this $Q\rightarrow \mathcal{H}_0$ is same as that of the unit element $\mathcal{H}_1\rightarrow \mathcal{H}_0$ in $B\mathcal{H}$. This is what I meant to ask. I am editing the question so that it makes some sense. @BertramArnold – Praphulla Koushik Dec 4 at 13:13
  • As you said a general morphism of stacks can't preserve "unit" principal bundles. If both stacks are presented by groupoids and the morphism is presented by a functor it sends the unit bundle of $\mathcal G$ to the pullback of the unit bundle of $\mathcal H$ along the object map of the functor (which should be something like the formula at the end of my previous comment), so in case of a Lie groupoid extension it does preserve unit bundles. All of this should more or less follow from the definitions, in particular how a Lie functor defines a fibered functor between the presented stacks. – Bertram Arnold Dec 4 at 13:48
up vote 3 down vote accepted

This is more of a long comment, but since the question has been answered in the last edits I will post it as an answer.

There are multiple perspectives on the stack presented by a Lie groupoid $\mathcal G$: One is the fibered category of principal bundles which you use, another one is the sheafification of the presheaf of groupoids which sends $X$ to the groupoid with objects maps from $X$ to $\mathcal G_0$ and morphisms maps from $X$ to $\mathcal G_1$. These correspond to trivial principal bundles; in the stackification we have to replace our test manifold by a cover $U$ and consider trivial principal bundles on $U$ together with descent data on $U\times_X U$. You can check that these glue to a principal bundle on $X$. (If $\mathcal G = pt//G$, this is just the description of principal $G$-bundles by cocycles.)

Both descriptions are equivalent, and the principal bundle description is arguably nicer - it involves a short list of data and relations, whereas the stackification of the prestack requires you to allow all possible covers, then identify descent data on different covers when there is a common refinement, ... However this description only works well if we map into the stack $B\mathcal G$ (for instance, when evaluating it on a test manifold). When we map out of it, we usually use the cover coming from the Lie groupoid presentation:

If $\mathcal G$ and $\mathcal H$ are two Lie groupoids, one can ask what the groupoid of natural transformations between $B\mathcal G$ and $B\mathcal H$ is. By the Yoneda lemma, this should correspond to the evaluation of $B\mathcal H$ on $B\mathcal G$, and since the unit principal bundle defines a morphism $\mathcal G_0\to B\mathcal G$ which is a cover, this evaluation is just given by objects in $B\mathcal H(\mathcal G_0)$ together with an isomorphism between the two ways of pulling this object back to $\mathcal G_1$. You can check that this recovers the notion of a bibundle between Lie groupoids - the object in $B\mathcal H(\mathcal G_0)$ defines the right $\mathcal H$-bundle structure, and the isomorphism defines the left $\mathcal G$-bundle structure. From this description you can immediately derive all formulas for bibundles, e.g. the composition of bibundles, the bibundle associated to a functor, ...

Lastly, I think that while it's important to have a rigourous mathematical framework for stacks, for which fibered categories are certainly a good candidate, it's also important to have intuition about them, and for this I usually pretend that my stack is $BG$ when I map into it and the Cech groupoid of a cover of some manifold when I map out of it.

  • When you have a map of stacks $\mathcal{D}\rightarrow \mathcal{C}$ you want to think $\mathcal{C}$ as stack associated to a manifold $M$, some denote this by $\underline{M}=B\{M\rightrightarrows M\}$ and you want to think $\mathcal{D}$ as stack associated to a Lie groupoid $\mathcal{G}$ which is denoted by $B\mathcal{G}$... Is this what you mean in last paragraph? – Praphulla Koushik Dec 4 at 17:30
  • " If $\mathcal G$ and $\mathcal H$ are two Lie groupoids, one can ask what the groupoid of natural transformations between $B\mathcal G$ and $B\mathcal H$ is. By the Yoneda lemma, this should correspond to the evaluation of $B\mathcal H$ on $B\mathcal G$" - Wait.. this would be a 2-version of Yoneda I guess? And by the way what is $B\mathcal{H}(B\mathcal{G})$ (if not tautologically the Hom groupoid $\mathrm{Stacks}(B\mathcal{G},B\mathcal{H})$ of $1$-morhpisms (="Lie functors") from $B\mathcal{G}$ to $\mathcal{H}$)? – Qfwfq Dec 4 at 19:28
  • @Qfwfq Even I do not understand what is "evaluation of $B\mathcal{H}$ on $B\mathcal{G}$"... The $2$-version of Yoneda lemma I am aware of is (which you can find for example in page $23$ of Orbifolds as Stacks by Eugene Lerman) that, "given a manifold $M$ and a Lie groupoid $\mathcal{G}$, the category of maps/natural transformations from $\underline{M}$ to $B\mathcal{G}$ is evaluation of $B\mathcal{G}$ on $M$" i.e., the fiber $B\mathcal{G}(M)$ which are just principal $\mathcal{G}$ bundles over $M$.. I do not know if this can be generalized to the case $B\mathcal{G}\rightarrow B\mathcal{H}$.. – Praphulla Koushik Dec 4 at 19:51
  • @Qfwfq appropriate thing on other side would be "Evaluation of $B\mathcal{H}$ on $B\mathcal{G}$" just like in case of $\underline{M}\rightarrow B\mathcal{G}$ it was evaluation of $B\mathcal{G}$ on $M$... What I know is there is a correspondence between maps of the form $B\mathcal{G}\rightarrow B\mathcal{H}$ and what are known as $\mathcal{G}-\mathcal{H}$ bibundles.. – Praphulla Koushik Dec 4 at 19:53
  • @PK: yep, that one you quote may indeed be called a "2-Yoneda". I was talking about a (hypothetical) version with objects of 2-categories (instead of objects of 1-categories) everywhere. – Qfwfq Dec 4 at 20:22

For any one wondering how did I write that

$(\mathcal{G}_1\times_{\mathcal{G}_0}(\mathcal{G}_0\times_{\mathcal{H}_0}\mathcal{H}_1))/\mathcal{G}_1$ is same thing as $\mathcal{G}_0\times_{\mathcal{H}_0}\mathcal{H}_1$,

here is the map between these two manifolds.

Given $(g,(x,h))\in \mathcal{G}_1\times_{\mathcal{G}_0}(\mathcal{G}_0\times_{\mathcal{H}_0}\mathcal{H}_1)$ we have $s(g)=x$ and $\phi(x)=t(h)$.

Given such element we have to associate an element $(y,l)$ in $\mathcal{G}_0\times_{\mathcal{H}_0}\mathcal{H}_1$ i.e., $y\in \mathcal{G}_0$ and $l\in \mathcal{H}_1$ such that $\phi(y)=t(l)$.

As $s(g)=x$ and $\phi(x)=t(h)$ we have $\phi(s(g))=t(h)$. This suggest to declare $y=s(g)$ and $l=h$. But then, it turns out that this does not give a well defined map from the quotient $(\mathcal{G}_1\times_{\mathcal{G}_0}(\mathcal{G}_0\times_{\mathcal{H}_0}\mathcal{H}_1))/\mathcal{G}_1$.

We have $\phi(s(g))=t(h)$ i.e., $s(\phi(g))=t(h)$ i.e., $\phi(g)h$ makes sense in $\mathcal{H}_1$.

We have $\phi(t(g))=t(\phi(g))=t(\phi(g)h)$. This suggests to declare $y=t(g)$ and $l=\phi(g)h$.

So, we have a map $\mathcal{G}_1\times_{\mathcal{G}_0}(\mathcal{G}_0\times_{\mathcal{H}_0}\mathcal{H}_1)\rightarrow \mathcal{G}_0\times_{\mathcal{H}_0}\mathcal{H}_1$, given by $$(g,(x,h))\mapsto (t(g),\phi(g)h)$$ It turns out that this map is well defined on quotient i.e., if $(g,(x,h))\sim (g',(x',h'))$ then, $t(g)=t(g')$ and $\phi(g)h=\phi(g')h'$. This defines a map $$(\mathcal{G}_1\times_{\mathcal{G}_0}(\mathcal{G}_0\times_{\mathcal{H}_0}\mathcal{H}_1))/\mathcal{G}_1\rightarrow \mathcal{G}_0\times_{\mathcal{H}_0}\mathcal{H}_1$$ given by $$(g,(x,h))\mapsto (t(g),\phi(g)h)$$ It turns out that this is a diffeomorphism. Thus,

$(\mathcal{G}_1\times_{\mathcal{G}_0}(\mathcal{G}_0\times_{\mathcal{H}_0}\mathcal{H}_1))/\mathcal{G}_1$ is same thing as $\mathcal{G}_0\times_{\mathcal{H}_0}\mathcal{H}_1$.

So, the principal $\mathcal{H}$ bundle $(\mathcal{G}_1\times_{\mathcal{G}_0}(\mathcal{G}_0\times_{\mathcal{H}_0}\mathcal{H}_1))/\mathcal{G}_1\rightarrow \mathcal{G}_0$ is same thing as principal $\mathcal{H}$ bundle $\mathcal{G}_0\times_{\mathcal{H}_0}\mathcal{H}_1\rightarrow \mathcal{G}_0$.

Now, when we are looking at morphisms $\phi$ such that $\phi:\mathcal{G}_0\rightarrow \mathcal{H}_0$ is the identity map (which happens when we are dealing with Lie groupoid extensions) the manifold $\mathcal{G}_0\times_{\mathcal{H}_0}\mathcal{H}_1$ is nothing but $\mathcal{H}_1$ and the principal $\mathcal{H}$ bundle $\mathcal{G}_0\times_{\mathcal{H}_0}\mathcal{H}_1\rightarrow \mathcal{G}_0$ is nothing but the target map $t:\mathcal{H}_1\rightarrow \mathcal{H}_0$.

So, we have following result.

Given a morphism of Lie groupoids $\phi:\mathcal{G}\rightarrow \mathcal{H}$, the corresponding map of stacks $F:B\mathcal{G}\rightarrow B\mathcal{H}$ preserves unit element upto a pull back in the sense that $F(t:\mathcal{G}_1\rightarrow \mathcal{G}_0)$ is pull back of principal $\mathcal{H}$ bundle $t:\mathcal{H}_1\rightarrow \mathcal{H}_0$ under the map $\phi:\mathcal{G}_0\rightarrow \mathcal{H}_0$.

As a special case when $\phi:\mathcal{G}_0\rightarrow \mathcal{H}_0$ is identity map, $F(t:\mathcal{G}_1\rightarrow \mathcal{G}_0)$ is equal to $t:\mathcal{H}_1\rightarrow \mathcal{H}_0$.

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