I am reading "Some notes on Differentiable stacks" by J. Heinloth. In that paper, the notion of quotient stack is defined as follows.

Let $G$ be a Lie group action on a manifold $X$ (left action). We define the quotient stack $[X/G]$ as $[X/G](Y):=\{P\xrightarrow{p} Y, P\xrightarrow{f}X | P\rightarrow Y \text{ is a G-bundle,} ~ f \text{ is } G\text{-equivariant}\}$.

Morphisms of objects are $G$-equivariant isomorphisms.

I am trying to understand the motivation for defining in this way.

Given a Lie group action of $G$ on $X$, if I want to associate a stack, I would start with simpler cases which allows me to guess how to define.

  • Given a Lie group $G$, I have stack associated to it, denoted by $BG$, the stack of principal $G$ bundles.

  • Given a manifold $M$, I have the stack associated to it, denoted by $\underline{M}$ whose objects are maps $Y\rightarrow M$.

Suppose $X$ is trivial and $G$ acts trivially on $X=\{*\}$ then $[X/G]$ should only depend on $G$. We know what stack to associate for a Lie group $G$ i.e., $BG$. Thus, $[X/G]$ should just be $BG$.

Suppose $G$ is trivial and $G$ acts on $X$, $[X/G]$ should only depend on $X$. We know what stack to associate for a manifold $X$ i.e., $\underline{X}$. Thus, $[X/G]$ should just be $\underline{X}$.

Suppose $G$ is non trivial and $X$ is non trivial and that the action of $G$ on $X$ is such that $X/G$ is a manifold. We know what stack to associate for a manifold $X/G$ i.e., $\underline{X/G}$. Thus, $[X/G]$ should just be $\underline{X/G}$.

I am not able to guess how could we guess the definition knowing above three cases. Is quotient stack definition motivated from these simpler cases or Is it the case that simpler cases are special cases of notion of Quotient stack. How could we come up with such a definition. Any comments regarding the motivation are welcome.

  • $\begingroup$ Your third case is wrong. You also need the action of $G$ to be free on $X$ (else you lose the stabilizers!) $\endgroup$ Dec 19, 2018 at 10:40
  • $\begingroup$ @DenisNardin When I said $G$ acts on $X$ so that $X/G$ is manifold I had in my mind quotient manifold theorem.. There $G$ acts on $X$ freely, properly (smoothly of course)... It then says $X/G$ is a manifold :D I did not even think about the case where $G$ action is not proper but $X/G$ is still a manifold... I should have mentioned free,, $\endgroup$ Dec 19, 2018 at 10:43
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    $\begingroup$ You seem to forget the elementary fact that, since $f$ is $G$-equivariant, it simply induces a quotient map $P/G=Y\rightarrow X/G$. Hence if $X/G$ really existed as a manifold then the quotient stack $[X/G]$ would really be given by $X/G$. Else, we take this as a definition for $[X/G]$ and it works fine, as explained in the answers below. That's how I understand it. $\endgroup$ Dec 19, 2018 at 13:39
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    $\begingroup$ @Louis-ClémentLEFÈVRE While what you say clearly proves that there is a map $[X/G]\to X/G$, I think it is a bit misleading: take $X=*$. Then $*/G$ exists as a manifold (it's just $*$ again) but it's certainly not the same as $[X/G]=\mathbf{B}G$. You do need some more hypotheses to deduce that the map $[X/G]\to X/G$ is an isomorphism $\endgroup$ Dec 19, 2018 at 13:46
  • $\begingroup$ @Louis-ClémentLEFÈVRE We assumed $X/G$ is a manifold.... If $f:P\rightarrow X$ is $G$-equivariant, this induces map $P/G\rightarrow X/G$.. Ok... As $P\rightarrow Y$ is principal $G$-bundle, $P/G=Y$... Ok till here... Thus, $f:P\rightarrow X$ give a map $Y\rightarrow X/G$.... Ok ... So, we have a map of categories $[X/G]\rightarrow \underline{X/G}$... Under good conditions, this map is an isomorphism... This is what you are saying... Right? $\endgroup$ Dec 19, 2018 at 17:55

4 Answers 4


Let's start approaching the question from the simplest possible case $Y=*$. What should be the points of $[X/G]$?

Recall that the idea here is to generalize the construction of the action groupoid for discrete groups acting on sets to the manifold case. This allows us to remember the stabilizers of points and it is generally a much better behaved notion.

So morally $[X/G](*)$ should be the groupoid whose objects are points of $X$ and such that $\mathrm{Mor}(x,x')=\{g\in G\mid gx=x'\}$. With this description however it is a bit unclear how to generalize that to get a description of $[X/G](Y)$, so let us rewrite it in a slightly different way.

A point of $X$ is just a $G$-equivariant morphism $G\to X$ (since any such $G$-equivariant morphism is determined by the image of $e\in G$). Moreover a morphism between $x:G\to X$ and $x':G\to X$ is exactly a $G$-equivariant morphism $g:G\to G$ (i.e. right multiplication by some element $g\in G$) making the obvious diagram commute. Now if you look at the definition, the groupoid does not depend from the fact that $G$ has a canonical basepoint (the identity element $e\in G$), so in fact we can write

$[X/G](*)$ is defined as the groupoid of $G$-equivariant maps $T\to X$ where $T$ is a freely transitive $G$-space.

Ok, so now we want to describe the groupoid $[X/G](Y)$. Intuitively the objects here should be families of elements of $[X/G](*)$ parametrized by $Y$. But a family of freely transitive $G$-spaces is exactly a principal $G$-bundle $P\to Y$, and a family of $G$-equivariant map $P_y\to X$ for each $y\in Y$ is just a $G$-equivariant map $P\to X$. Hence we get the definition you are asking about.

  • $\begingroup$ By points of $[X/G]$ you mean objects of $[X/G](*)$.. right? I am feeling shy to say I do not understand "naturally its objects should be $G$-equivariant maps $G\rightarrow X$".... But I have no option... :D Can you please tell why you think it is natural... $\endgroup$ Dec 19, 2018 at 10:48
  • $\begingroup$ Well, the idea of the quotient stack is to replace $X/G$ by its action groupoid (to remember the all important stabilizers!). The action groupoid on a set $X$ is exactly the groupoid of $G$-equivariant maps from a freely transitive $G$-set $T$ to $X$. Or if you want, the groupoid of points of $X$ where maps are elements of $G$ sending one point to the next, but my first description lends itself more easily to treating the case of families $\endgroup$ Dec 19, 2018 at 10:50
  • $\begingroup$ Oh Oh Oh... It makes so much sense... given a Lie group action $G$ on $X$, I have a Lie groupoid associated to it (If I remember correctly I saw the notation $X/G$ for this)... For this Lie groupoid $\mathcal{G}$, I have the notion of stack $B\mathcal{G}$ the stack of principal groupoid $\mathcal{G}$ bundles.. I declare $[X/G]$ to be $B\mathcal{G}$.. This makes so much sense.. I now feel like an Idiot :D :D $\endgroup$ Dec 19, 2018 at 10:56
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    $\begingroup$ I expanded the answer to explain that this is just a generalization of the notion of "action groupoid" to families. $\endgroup$ Dec 19, 2018 at 10:57
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    $\begingroup$ A comment: If you take the naive definition of the functor of points for all Y as an action groupoid, you will get something which doesn't satisfy descent - a presheaf /prestack so to say. If you then sheafify it (in the appropriate stacky /homotopy sense), you recover the definition in terms of principal bundles. $\endgroup$
    – Asvin
    Dec 19, 2018 at 16:41

Let me detail the following case: $G$ is a Lie group that acts freely and properly on the manifold $X$. Classical theorems in differential geometry state that the quotient is a manifold and that $X\rightarrow X/G$ is a principal $G$-bundle. Then as I said in the comments, given a principal $G$-bundle $P\rightarrow Y$ and a $G$-equivariant map $f:P\rightarrow X$ then at the quotient $f$ induces a map $g:Y\rightarrow X/G$, simply because $P/G=Y$.

Conversely, again in this situation for the action of $G$, let's take a map of manifolds $g:Y\rightarrow X/G$. Since the bundle $X\rightarrow X/G$ is locally trivial one can lift locally on $Y$ the map $g$ to a map $U\times G \rightarrow X$ (with $U\subset Y$) which is $G$-equivariant, seeing $U\times G$ as a trivial principal $G$-bundle. Such a lift is not unique: you can change it by an automorphism of the principal $G$-bundle $U\times G$. Glueing these lifts defines a principal $G$-bundle $P$ over $Y$ together with a $G$-equivariant map $P\rightarrow X$ lifting $g$. This corresponds exactly to the descent theory for principal $G$-bundles.

Let me add also that the theory of stacks is made so that, without any assumption on the action of $G$, $X\rightarrow [X/G]$ should be a principal $G$-bundle. More precisely, this means that for any map $Y\rightarrow [X/G]$, the fiber product in the category of stacks $Y\times_{[X/G]} X$ is in fact a manifold $P$ and that the projection $P\rightarrow Y$ is a principal $G$-bundle (in other words the principal $G$-bundle $X\rightarrow [X/G]$ pulls back to a principal $G$-bundle $P\rightarrow Y$ in the usual sense). This is the standard way of defining geometric conditions for stacks and morphisms between them: by pulling it back along a map from a manifold. This forces you again to take this definition for the stack $[X/G]$.

  • $\begingroup$ Thanks for your answer... First paragraph is precisely what I have said in comments.. We assume the action of $G$ on $X$ is free and proper so that $X/G$ is a manifold.... With the definition of $[X/G]$ as mentioned in my question, given a manifold $Y$ and an object in $[X/G](Y)$ we get an object in $\underline{X/G}(Y)$ i.e., a map $Y\rightarrow X/G$.. This gives a functor $[X/G]\rightarrow \underline{X/G}$... (continued) $\endgroup$ Dec 20, 2018 at 13:12
  • $\begingroup$ Conversely, suppose we are given a map $Y\rightarrow X/G$... We need a manifold $P$ and maps $P\rightarrow Y$ and $P\rightarrow X$ so that $P\rightarrow Y$ is a principal $G$ bundle and $P\rightarrow X$ is $G$-equivariant... As $G$ action is proper and free, $X\rightarrow X/G$ is principal $G$ bundle.. We pull back this along $Y\rightarrow X/G$ to get a principal $G$ bundle $P\rightarrow Y$.. We actually have $P=Y\times_{X/G}X$.... This $P$ comes with obvious projection map $P\rightarrow X$ with $(y,x)\mapsto x$... Thus, we get an element of $[X/G](Y)$.. (continued) $\endgroup$ Dec 20, 2018 at 13:17
  • $\begingroup$ It should not be difficult to check that these gives an isomorphism $[X/G]\rightarrow \underline{X/G}$.... This is your second paragraph... I like the sentence "without any assumption on the action of $G$, $X\rightarrow [X/G]$ should be a principal $G$-bundle".. I did not understand at all when I read it for the first time... It makes so much sense now... That property is usally goes by the name representable morphism... (continued) $\endgroup$ Dec 20, 2018 at 13:26
  • $\begingroup$ A map of stacks $\mathcal{D}\rightarrow \mathcal{C}$ is representable plus some property (P) if, given any manifold $M$, the fiber product $\mathcal{D}\times_{\mathcal{C}}M$ is a stack that is a manifold and the projection map $\mathcal{D}\times_{\mathcal{C}}M\rightarrow M$ (at the level of manifolds) has the property (P).. Here the property is being principal $G$ bundle.. This is what you mean when you say map of stacks $X\rightarrow [X/G]$ is a principal $G$ bundle... This makes so much sense... (continued) $\endgroup$ Dec 20, 2018 at 13:28
  • $\begingroup$ By "This forces you again to take this definition for the stack $[X/G]$" you mean like some universal property?? You are saying if we look for a stack $\mathcal{D}$ so that $X\rightarrow \mathcal{D}$ is a principal $G$ bundle and if it is minimal (in some sense) then we end up $\mathcal{D}$ to be that $[X/G]$ as defined in the question... This is correct? $\endgroup$ Dec 20, 2018 at 13:37

The nicest possible action of $G$ on $X$ is the free one, because in this case the quotient $X/G$ is a manifold or not far from a manifold, the map $X\to X/G$ is a $G$-principal bundle, and $G$-equivariant geometry on $X$ is the same as geometry on $X/G$. In short, the situation is as nice as possible.

The construction of the quotient stack $[X/G]$ is motivated by the desire to obtain such good properties in the non-free case also. Let me try to show how the definition of $[X/G](Y)$ comes naturally, say in the case that $Y$ is a point for simplicity. For this, we make the following observation. Fundamentally the quotient of a space by a group action is the set of orbits. Now if you think about it, an orbit is nothing else than the image of an equivariant map $f:P\to X$ where $P$ is a principal homogeneous space (a.k.a. a $G$-bundle), and we have a free orbit exactly when $f$ is injective. What is special with the case when $G$ acts freely on $X$ is that in this case any map $f$ as above has to be injective. But if you change slightly your conception of an orbit and if instead of the image $f(P)$, you focus on $P$ itself (understood with its mapping to $X$), then you somehow restore all good properties of free orbits — indeed, after all the action on $P$ is free! Therefore if we call orbit an equivariant map from a $G$-bundle to $X$, then $[X/G]$ is just the set of orbits. Now deriving the definition for fibrations over a general space $Y$ is easy.

  • $\begingroup$ you mean not far from a manifold... I do not know what is a $G$-equivariant geometry of a manifold.. can you please point me to some references... $\endgroup$ Dec 21, 2018 at 9:13
  • $\begingroup$ Consider all sorts of geometric objects that you like to relate to your manifold $X$: open submanifolds, closed submanifolds, vector bundles on $X$, differential forms... If $X$ has an action on $G$, there are equivariant analogues: open or closed invariant submanifolds, $G$-equivariant vector bundles (i.e. spaces $V$ with $G$ action and an equivariant map $V\to X$ which is a vector bundle), equivariant differential forms... Etc, etc. I do not think that there is a reference dedicated especially to this concept. $\endgroup$ Dec 21, 2018 at 9:44
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    $\begingroup$ lmgtfy.com/?q=G-equivariant :-) $\endgroup$
    – David Roberts
    Dec 22, 2018 at 4:19
  • $\begingroup$ @DavidRoberts I know what is $G$-equivariant mean in sense of a map... I did not knew what is $G$-equivariant geometry. Thanks for your link. $\endgroup$ Dec 26, 2018 at 15:17

Given a Lie group action $G$ on $X$ we have what is called a action groupoid (Translation groupoid) associated with the action usually denoted by $G\ltimes X$.

Objects of this category are elements of $X$. Given $x,y\in (G\ltimes X)_0=X$, there is an arrow $x\rightarrow y$ if these two elements $x,y$ are related by an element of $G$ i.e., if there exists $g\in G$ such that that $g.x=y$. We can also write $$\text{Mor} (x,y)=\{g\in G:g.x=y\}$$

So, we have a Lie groupoid $\mathcal{G}=G\ltimes X$ associated with this action $G$ on $X$.

There is a notion of what is called a principal $\mathcal{G}$ bundle for a groupoid $\mathcal{G}$.

I mentioned that, for each Lie group $G$ there is a stack $BG$, for each manifold $M$ there is a stack $\underline{M}$.

I somehow forgot to mention that, for each Lie groupoid $\mathcal{G}$ there is a stack associated namely $B\mathcal{G}$, the category of principal $\mathcal{G}$ bundles.

A principal $\mathcal{G}$ bundle over a manifold $M$ is a surjective submersion $\pi:P\rightarrow M$ with an action of $\mathcal{G}$ on $P$ (which comes with a map $a:P\rightarrow \mathcal{G}_0$) such that

  • $\pi:P\rightarrow M$ is $\mathcal{G}$ invariant
  • the map $P\times_{\mathcal{G}_0}\mathcal{G}_1\rightarrow P\times_MP$ given by $(p,g)\mapsto g.p$ is an diffeomorphism.

This definition can be found online and in particular in the paper Orbifolds as Stacks by Eugene Lerman.

The category of these principal $\mathcal{G}$ bundles is denoted by $B\mathcal{G}$. This is a stack associated to $\mathcal{G}$.

Now, consider the case when $\mathcal{G}=G\ltimes X$. Object space is $X$ here. So, principal bundle comes with map $P\rightarrow X$. As mentioned above, it comes with map $P\rightarrow M$ which is $\mathcal{G}$ invarinat. As $\mathcal{G}_1=G$, this simply mean $P\rightarrow M$ is a principal $G$ bundle (neglecting some technicalities).

Thus, given a manifold $M$, the category $B\mathcal{G}(M)$ is a collection of principal $\mathcal{G}$ bundles over the base $M$. Thus objects are principal $\mathcal{G}$ bundles over $M$ which comes with map $P\rightarrow \mathcal{G}_0$ (anchor map) and $P\rightarrow M$ (projection map). In case when $\mathcal{G}=G\ltimes X$, this comes with map $P\rightarrow X$ and $P\rightarrow M$ such that $P\rightarrow M$ is $\mathcal{G}$ bundle and $P\rightarrow X$ is $\mathcal{G}$-equivariant.

$B\mathcal{G}(M):=\{P\xrightarrow{p} M, P\xrightarrow{f}M | P\rightarrow M \text{ is a } \mathcal{G}-\text{bundle,} ~ f \text{ is } \mathcal{G}\text{-equivariant}\}$.

This is precisely the definition for quotient stack.

$[X/G](Y):=\{P\xrightarrow{p} Y, P\xrightarrow{f}X | P\rightarrow Y \text{ is a G-bundle,} ~ f \text{ is } G\text{-equivariant}\}$.

This is (may be) the (one of) motivations for the definition of Quotient stack.

  • If action of the Lie group $G$ on the manifold $X$ is free and proper, what we get is a manifold $X/G$. Stack associated to this manifold is $\underline{X/G}$ which we call to be the quotient stack, denote by $[X/G]$.

  • If the action of the Lie group $G$ on the manifold $X$ is not necessarily free and proper, what we get is a Lie groupoid denoted (among other symbols) by $X//G$. Stack associated to this Lie groupoid $X//G$ is $B(X//G)$ which we call to be the quotient stack, denote by $[X/G]$.

  • $\begingroup$ What is the reason for downvote? $\endgroup$ Dec 19, 2018 at 17:08
  • $\begingroup$ I think it's more instructive to just with out what I mentioned in your own. It's only a matter of chasing down the right definitions and I found it to be an useful exercise. Maybe I can write something more detailed later if i have time but i would strongly suggest working it out on your own! $\endgroup$
    – Asvin
    Dec 19, 2018 at 17:16
  • $\begingroup$ I do not understand "I think it's more instructive to just with out what I mentioned in your own." I am thinking I have mentioned details.. If you think I am missing something, please let me know I will write in detail.. $\endgroup$ Dec 19, 2018 at 17:18
  • $\begingroup$ I mean think of the action groupoid as a prestack/presheaf and try and sheafify it. You will end up with G bundles $\endgroup$
    – Asvin
    Dec 19, 2018 at 17:19
  • $\begingroup$ "think of the action groupoid as a prestack/presheaf and try and sheafify it. You will end up with G bundles " Ok Ok.. I will do that... Thank you :) $\endgroup$ Dec 19, 2018 at 17:21

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