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Definition : Let $\mathcal{G}$ and $\mathcal{H}$ be Lie groupoids. A bibundle from $\mathcal{G}$ to $\mathcal{H}$ is a manifold $P$ together with two maps $a_L:P\rightarrow \mathcal{G}_0,a_R:P\rightarrow \mathcal{H}_0$ 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.

I am trying to understand in what sense these are called generalized morphisms between Lie groupoids.

There is already a notion of generalized morphsim between Lie groupoids from Ieke Moerdijk's article Orbifolds as groupoids.

Definition : A generalized morphism from a Lie groupoid $\mathcal{G}$ to a Lie groupoid $\mathcal{H}$ is a morphism of Lie groupoids $\mathcal{G}'\rightarrow \mathcal{H}$ where $\mathcal{G}'$ is a Lie groupoid morita equivalent to $\mathcal{G}$.

The name generalized morphisms seems reasonable for this but I do not see in what sense a bibundle is said to be a generalized morphism.

Any comments that helps to understand in what sense bibundles are called as generalized morphisms are welcome.

Is it that given a bibundle there is a generalised morphism in the sense I have defined above and given a generalised morphism there exists a bibundle associated to that? I could not find any reference.

I am also not very comfortable/happy with the point that $a_R$ is just $\mathcal{G}$ invariant. I was guessing that the condition would be $a_R$ is a principal $\mathcal{G}$ bundle just similar to the second condition where it says $a_L:P\rightarrow \mathcal{G}_0$ is a principal $\mathcal{H}$ bundle. Is there any reason why the definition do not ask $a_R$ to be principal $\mathcal{G}$ bundle?? What is the speciality of just $a_L$?

I would be very comfortable even if the definition is just the following :

Definition : Let $\mathcal{G}$ and $\mathcal{H}$ be Lie groupoids. A bibundle from $\mathcal{G}$ to $\mathcal{H}$ is a manifold $P$ together with two maps $a_L:P\rightarrow \mathcal{G}_0,a_R:P\rightarrow \mathcal{H}_0$ 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. $a_L$ is $\mathcal{H}$ invariant

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

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

To make sense of question of action of $\mathcal{G}$ and $\mathcal{H}$ being commuttaive, it is just sufficient to have $a_R(g.p)=a_R(p)$ and $a_L(p.h)=a_L(p)$. The definition of bibundle given is none of what I have expected. It is somewhere in between. What is the significance there?

Any comments are welcome.

https://ncatlab.org/nlab/show/bibundle says,

A bibundle is a groupoid principal bundle which is equipped with a compatible second groupoid action “from the other side”.

I am somehow ok with this definition. But the question of seeing bibundles as generalized morphisms is still not clear.

EDIT : It says in Page no $12$ of Orbifolds as stacks? that, to a $\mathcal{G}-\mathcal{H}$ bibundle $P$ one can associate a functor $\{\mathcal{G}-bundles\}\rightarrow \{\mathcal{H}-bundles\}$. Is this can be the reason for calling bibundles as generalized morphism where morphism of $\mathcal{G}$ to $\mathcal{H}$ is just a morphism of Lie groupoids $\mathcal{G}\rightarrow \mathcal{H}$ where as if you go one step above the ladder you have category of $\mathcal{G}$ bundles and your bibundle map is giving a morphism in that above step. Am I misunderstanding something here?

Just for convenience, I am attaching diagram

enter image description here

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    $\begingroup$ You use pullback to give a groupoid morphism between a pullback of $\mathcal G$ and $\mathcal H$. The role of the two actions are not symmetric because tour generalized morphism has a direction. If you ask both principality conditions you get an invertible generalized morphism, thus a Morita equivalence. $\endgroup$ – Nicola Ciccoli Jun 28 '18 at 8:38
  • $\begingroup$ @NicolaCiccoli Sir, It looks like I understand but I do not :D :D All I can say is that you mean suppose there is a bibundle $P:\mathcal{G}\rightarrow \mathcal{H}$ that is a principal $\mathcal{G}$ bundle then $\mathcal{G}$ and $\mathcal{H}$ are Morita equivalent. Am I correct? If that is the case, that would give sufficient justification for not considering bibundle that is principal on both sides.. Now, can you tell me what pull back you are talking here when you say pull back of $\mathcal{G}$ and $\mathcal{H}$? $\endgroup$ – Praphulla Koushik Jun 28 '18 at 9:41
  • $\begingroup$ @NicolaCiccoli Please let me know if you can say little more than that... $\endgroup$ – Praphulla Koushik Jun 29 '18 at 11:35
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As I mentioned in my other answer, Lie groupoids give rise to stacks, and the stack that a Lie groupoid $X$ gives rise to is the stack of principal $X$-bundles. Then in analogy with the Eilenberg-Watts theorem in the theory of rings, a map of such stacks is given by a bibundle. Eilenberg-Watts says that a suitable functor of module categories $Mod_R \to Mod_S$ for rings $R$ and $S$ is, up to isomorphism, nothing other than one whose effect on objects is tensoring an $R$-module with an $R$-$S$-bimodule (which is then an $S$-module). Likewise, a map from the stack of principal $X$-bundles to the stack of principal $Y$-bundles always arises by "tensoring" an $X$-bundle by an $X$-$Y$-bibundle. The definition is not quite so symmetric, since what you get out should be a principal bundle again, so you need the bibundle to be principal on one side, for the $Y$-action.

The short answer is that generalised morphisms should satisfy a universal property for bicategories, and taking bibundles as morphisms gives a bicategory of Lie groupoids that has that universal property. But this is somewhat unsatisfactory from the conceptual side.

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  • $\begingroup$ 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 12:24
  • $\begingroup$ I understand that given a Lie groupoid X there is a stack associated with it... Given Stacks BX and BY, a map of stacks $BX\rightarrow BY$ should come from a $X-Y$ bibundle. Are you saying this is why bibundles are called generalised morphisms??? Generalised morphisms are not morphisms of Lie groupoids but they are something coming from morphism of stacks coming from Lie groupoids.. Did I get it correctly?? $\endgroup$ – Praphulla Koushik Jul 6 '18 at 20:53
  • $\begingroup$ I do not understand “generalised morphisms should satisfy a universal property for bicategories, and taking bibundles as morphisms gives a bicategory of Lie groupoids that has that universal property. ” can you say little more.. I got answer for what I have asked.. this is something extra.. thanks for that... $\endgroup$ – Praphulla Koushik Jul 6 '18 at 20:55
  • $\begingroup$ @PraphullaKoushik I am travelling and will respond to your comments when possible. $\endgroup$ – David Roberts Jul 7 '18 at 23:26
  • $\begingroup$ No problem :) :) Thank you thank you $\endgroup$ – Praphulla Koushik Jul 8 '18 at 3:03

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