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 $\mathcal{H}$.

  • $\begingroup$ 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). $\endgroup$ Dec 4, 2018 at 11:53
  • $\begingroup$ I think you mean to say "[---] If you present the morphism as a functor $F:\mathcal{G}\rightarrow \mathcal{H}$ [---]"... @BertramArnold $\endgroup$ Dec 4, 2018 at 12:42
  • $\begingroup$ 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 $\phi:\mathcal{G}_1\rightarrow \mathcal{H}_1$). $\endgroup$ Dec 4, 2018 at 13:13
  • $\begingroup$ 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 $\endgroup$ Dec 4, 2018 at 13:13
  • $\begingroup$ 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. $\endgroup$ Dec 4, 2018 at 13:48

1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$ Dec 4, 2018 at 17:30
  • $\begingroup$ " 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 $\mathcal{H}$)? $\endgroup$
    – Qfwfq
    Dec 4, 2018 at 19:28
  • $\begingroup$ @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 $B\mathcal{G}\rightarrow B\mathcal{H}$.. $\endgroup$ Dec 4, 2018 at 19:51
  • $\begingroup$ @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.. $\endgroup$ Dec 4, 2018 at 19:53
  • $\begingroup$ @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. $\endgroup$
    – Qfwfq
    Dec 4, 2018 at 20:22

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.