I am reading Orbifolds as stacks?

Given Lie groupoids $\mathcal{G}$ and $\mathcal{H}$ there is a notion of what is called a bibundle from $\mathcal{G}$ to $\mathcal{H}$ which is supposed to be a ageneralized notion of a morphism of Lie groupoids.

(rough) Definition : A bibundle is a groupoid principal bundle which is equipped with a second groupoid action from the other side.

(precise) Definition : A bibundle from $\mathcal{G}$ to $\mathcal{H}$ is a manifold $P$ together with maps $a_L:P\rightarrow \mathcal{G}_0$ and $a_R:P\rightarrow \mathcal{H}_0$ such that

  1. there is a left action of $\mathcal{G}$ on $P$ with respect to an anchor $a_L$ and a right action of $\mathcal{H}$ on $P$ with respect to an anchor $a_R$.

  2. $a_L:P\rightarrow \mathcal{G}_0$ is a principal $H$-bundle.

  3. $a_R$ is $\mathcal{G}$ invariant.

  4. the actions of $\mathcal{G}$ and $\mathcal{H}$ commutes.

The point is to make collection of Lie groupoids as a $2$ category with Lie groupoids seen as objects, bibundles as morphisms between objects and isomorphisms of bibundles as maps between bibundles.

For that, we need to define what does it mean to say composition of two maps between Lie groupoids i.e., given bibundles $P:\mathcal{G}\rightarrow \mathcal{H}$ and $Q:\mathcal{H}\rightarrow \mathcal{K}$, we need to declare what is $P\circ Q:\mathcal{G}\rightarrow \mathcal{K}$.

We want it to be a bibundle. So, we need atleast maps $a_L:P\circ Q\rightarrow \mathcal{G}_0$ and $a_R:P\circ Q\rightarrow \mathcal{K}_0$.

What we have is $a_L:P\rightarrow \mathcal{G}_0,a_R:P\rightarrow \mathcal{H}_0,a_L:Q\rightarrow \mathcal{H}_0,a_R:Q\rightarrow \mathcal{K}_0$ such that both $a_L$ are principal bundles.

One natural thing (which I did not realize before) is to consider pullback $P\times_{\mathcal{H}_0}Q$ from maps $a_R:P\rightarrow \mathcal{H}_0,a_L:Q\rightarrow \mathcal{H}_0$ that gives maps $\pi_1:P\times_{\mathcal{H}_0}Q\rightarrow P$ and $\pi_2:P\times_{\mathcal{H}_0}Q\rightarrow Q$ which on composition with $a_L:P\rightarrow \mathcal{G}_0, a_R:Q\rightarrow \mathcal{K}_0$ gives $a_L\circ \pi_1:P\times_{\mathcal{H}_0}Q\rightarrow \mathcal{G}_0$ and $a_R\circ \pi_2:P\times_{\mathcal{H}_0}Q\rightarrow \mathcal{K}_0$. As $a_L$ is a submerison, $P\times_{\mathcal{H}_0}Q$ has a smooth manifold structure.

It feels like $P\times_{\mathcal{H}_0}Q$ is the candidate for composition. As said before, we atleast need a manifodl $P\circ Q$ and maps $P\circ Q\rightarrow \mathcal{G}_0$ and $P\circ Q\rightarrow \mathcal{K}_0$.

Along with this, we need action of $\mathcal{G}$ on $P\circ Q$ and an action of $\mathcal{H}$ on $P\circ Q$.

We have obvious action maps

  • $\mathcal{G}_1\times_{\mathcal{G}_0}(P\times_{\mathcal{H}_0}Q)\rightarrow P\times_{\mathcal{H}_0}Q$ with $(g,(p,q))\mapsto (g.p,q)$

  • $(P\times_{\mathcal{H}_0}Q)\times_{\mathcal{H}_0}\mathcal{H}_1\rightarrow P\times_{\mathcal{H}_0}Q$ with $((p,q),h)\mapsto (p,q.h)$

It looks this should be the composition but that paper says some thing more.

Since the action of $\mathcal{H}$ on $P$ is principal, the action of $\mathcal{H}$ on $P\times_{\mathcal{H}_0}Q$ given by $(p,q).h=(p.h,h^{-1}q)$ is free and proper. Hence the quotient $(P\times_{\mathcal{H}_0}Q)/\mathcal{H}$ is a manifold.

I am not able to understand what above statements mean. I have not come across the notion of quotient of a manifold by a Lie groupoid. I only know that if a Lie group acts properly, freely on a manifold, the quotient is a maniofld. What Lie group they are considering here? $\mathcal{H}_1$? I guess no, It is not even a group even though there is a smooth structure.

Can some one help me to see what I am missing here?

EDIT : I think I misunderstood that they are talking about a Lie group action on manifold $P\times_{\mathcal{H}_0}Q$ when they are saying $(P\times_{\mathcal{H}_0}Q)/\mathcal{H}_1$. Even though $\mathcal{H}_1$ is not a Lie group they can still consider the quotient space given by relation coming form the action of $\mathcal{H}_1$.

Let $(p_1,q_1),(p_2,q_2)\in P\times_{\mathcal{H}_0}Q$. We declare $(p_1,q_1)\sim(p_2,q_2)$ if there exists $h\in \mathcal{H}_1$ such that $p_1=p_2.h$ and $q_1=h^{-1}q_2$. This is an equivalence relation on $P\times_{\mathcal{H}_0}Q$. I guess they mean $(P\times_{\mathcal{H}_0}Q)/\sim$ in my notation when they say $(P\times_{\mathcal{H}_0}Q)/\mathcal{H}$.

If this is the case, then they use some theorem which says when a quotient space of a manifold given by a relation is a manifold to conclude $(P\times_{\mathcal{H}_0}Q)/\mathcal{H}$ i.e., $(P\times_{\mathcal{H}_0}Q)/\sim$ in my notation is a manifold.

Can some one please confirm if this is what it is.

EDI : STACKY LIE GROUPS says in page $6$ that

Let $\mathcal{G},\mathcal{H},\mathcal{K}$ be Lie groupoids and $M$ be a $\mathcal{G}-\mathcal{H}$ bibundle and $N$ be a $\mathcal{H}-\mathcal{K}$ bibundle. Viewing a bibundle as relation of stacks suggests defining the composition of bibundles as $M\circ N=(M\times_{\mathcal{H}_0}N)/\mathcal{H}_1$ where the quotient is with respect to diagonal action $(m,n).h=(m.h,h^{-1}.n)$ for $m,n$ wherever defined.

I am not aware of seeing bibundles as relation of stacks. Can some one help me to see bibundles as relation of stacks. May be then, It would be easier for me to understand why that choice would natural when defining composition.

  • $\begingroup$ Maybe it is a misprint - the action of $\mathcal H$ on $Q$ is principal, not on $P$. It does not actually matter for the rest I believe: if $\mathcal H$ acts freely on either one of $P$ or $Q$ then it also acts freely on $P\times_{\mathcal H_0}Q$. $\endgroup$ – მამუკა ჯიბლაძე Jun 27 '18 at 18:39
  • $\begingroup$ @მამუკაჯიბლაძე what does it mean to say quotient here? $\endgroup$ – Praphulla Koushik Jun 28 '18 at 0:54
  • $\begingroup$ My recommendation is not to use bibundles, but anafunctors, as defined by Makkai (and then Bartels), which turn out to be equivalent. This is covered in tac.mta.ca/tac/volumes/26/29/26-29abs.html (shameless plug). $\endgroup$ – David Roberts Jun 28 '18 at 6:53
  • 1
    $\begingroup$ The quotient is as you write in your last edit. One useful intuition might be this: $P$ induces a functor from right $\mathcal G$-bundles to right $\mathcal H$-bundles; in the same way $Q$ induces a functor from right $\mathcal H$-bundles to right $\mathcal K$-bundles. Then if you ask whether the composite of these functors is also induced in the same way by some bibundle, and if you compute what it can be, exactly the one in your question will come out. $\endgroup$ – მამუკა ჯიბლაძე Jun 28 '18 at 7:23
  • 2
    $\begingroup$ @PraphullaKoushik well, composition of anafunctors is merely (strict) pullback, rather than some complicated induced change of structure groupoid as for bibundles. You need to push the complication somewhere, and it goes into vertical composition of 2-arrows, but most of the time you really just want to worry about 1-arrows and their composition, which is easier for anafunctors. $\endgroup$ – David Roberts Jun 28 '18 at 12:10

I will give an example as to why composition of bibundles cannot simply be done as pullback, as well as the relation to perhaps more familiar geometric constructions.

Firstly, note that if $M$ is a manifold and $\mathbf{B}G$ is a one-object Lie groupoid (where the automorphisms of the one object are the Lie group $G$), a bibundle from $M$ to $\mathbf{B}G$ is precisely the same thing as a principal $G$-bundle on $M$. Now consider a homomorphism $\phi\colon G\to H$, and the corresponding functor $\mathbf{B}G\to \mathbf{B}H$. This gives a bibundle from $\mathbf{B}G$ to $\mathbf{B}H$, namely $u(H) \to \ast = Obj(\mathbf{B}G)$, where $u(H)$ is the underlying manifold of the Lie group $H$. The left $\mathbf{B}G$-action is trivial, and the right $\mathbf{B}H$-action (which is the same as an $H$-action) is by multiplication. The naive "composition" of bibundles should give a bibundle from $M$ to $\mathbf{B}H$, which is just a principal $H$-bundle on $M$, but what it actually gives is just $P\times H \to M$, which is the composite of the projection on $P$ and the given map $P\to M$.

The correct composite should be the principal bundle you get by changing the structure group along the given homomorphism $G\to H$, which is the quotient $(P\times H)/G$ by the action of $G$ on $P\times H$ as $(p,h) \mapsto (pg,\phi(g)^{-1}h)$. Equivalently, one can set up an equivalence relation on $P\times H$ so that $(pg,h) \simeq (p,\phi(g)h)$. This generalises directly to the case when you replace $\mathbf{B}G\to \mathbf{B}H$ by some arbitrary functor $X\to Y$ of Lie groupoids (and compose with the bibundle it gives rise to), and with only a little more work when you replace the manifold $M$ by a general Lie groupoid. At that point, you probably will be comfortable with the general case.

  • $\begingroup$ Hello, I am with so much pending work :( I will respond in 2/3 days. Apologies. Thanks for your answer. I am sure it will make a difference :) $\endgroup$ – Praphulla Koushik Jul 4 '18 at 12:23
  • $\begingroup$ I am having little difficulty with the notations.. When you say “BG is one-object Lie groupoid” you mean $G$ is a Lie group, we have a groupoid associated to a Lie group, object set is Singleton, morphism set is G, usually written as $\{G\rightrightarrows *\}$.. you are calling this associated Lie groupoid also by G and then you are considering stack associated to Lie groupoid G(in your notation) and calling it BG. Is this correct? Or are you calling Lie groupoid it self by $BG$? $\endgroup$ – Praphulla Koushik Jul 6 '18 at 20:17
  • $\begingroup$ The notation $\mathbf{B}G$ means the Lie groupoid whose object manifold is $\{pt\}$ and whose arrow manifold is denoted $G$. It follows from the axioms that $G$ is nothing else than a Lie group. If I were to write the stack of $G$-bundles, which is the stackification of this Lie groupoid, I would write it as $\mathrm{Bun}_G$ or similar. $\endgroup$ – David Roberts Jul 15 '18 at 12:46
  • $\begingroup$ Can you check my other replies to this question as answers. YOur comments would be helpful.. $\endgroup$ – Praphulla Koushik Sep 26 '18 at 2:40
  • $\begingroup$ I'm not sure what else I would say. I prefer my counterexample to your counterexample, as it is shorter and less confusing to me. $\endgroup$ – David Roberts Sep 26 '18 at 4:09

I am just trying to rewrite what David Roberts have written in a notation that is more convinient for me. I am used to the notation $B\mathcal{G}$ to mean stack of Principal $\mathcal{G}$ bundles. So, even if you write $BG$ and not $B\mathcal{G}$, I am thinking it as a stack in my head so that is creating a problem.

Let $M$ be a manifold and $\{M\rightrightarrows M\}$ be the corresponding Lie groupoid. Let $G$ be a Lie group and $\{G\rightrightarrows *\}$ be the corresponding Lie groupoid. Let $P$ be a $\{M\rightarrow M\}-\{G\rightrightarrows *\}$ bibundle. As mentioned somewhere, a $\mathcal{G}-\mathcal{H}$ bibundle can be visualized as following diagram enter image description here

So, our $\{M\rightarrow M\}-\{G\rightrightarrows *\}$ bibundle is seen as following diagram enter image description here

which says that it just means $P\rightarrow M$ is a principal $G$ bundle ($G$ is a Lie group here).

Let $\phi:G\rightarrow H$ be a morphism of Lie groups. This gives a $\{G\rightrightarrows *\}-\{H\rightrightarrows *\}$ bibundle

enter image description here

So, writing these two bibundles in one diagram, we have

enter image description here

We have a principal $G$ bundle $P\rightarrow M$ and we want to associate a principal $H$ bundle over $M$. We use the construction of associated bundle. Any morphism of Lie groups $f:G\rightarrow H$ gives an action of $G$ on $H$ as $G\times H\rightarrow H$ with $(g,h)\mapsto f(g)^{-1}h$. We can then consider the associated bundle corresponding to this action of $G$ on $H$. We get a fiber bundle with base space $M$ and total space $(P\times H)/G$ with fiber $H$. It turns out that this fiber bundle is a principal $H$ bundle. We thus have following diagram

enter image description here

The map $(P\times H)/G\rightarrow M$ is given by $[p,h]\mapsto \pi(p)$ which is well defined.

For a more general notion of morphism of Lie groupoids, $(\mathcal{G}_1\rightrightarrows \mathcal{G}_0)\rightarrow (\mathcal{H}_1\rightrightarrows \mathcal{H}_0)$ and corresponding $\mathcal{G}-\mathcal{H}$ bibundle $Q$, it is only natural to consider the relative product $P\times_{\mathcal{H}_0}Q$ and then the quotient $(P\times_{\mathcal{H}_0})/\mathcal{G}_1$ just like in the case of morphism of Lie groups.

We have following diagram enter image description here

Thus, it is only natural to think of this definition of composition of bibundles.


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.