What are Lie groupoids intuitively?

MathOverflow Asked by Praphulla Koushik on September 30, 2020

I am trying to understand about Lie groupoids but not able to get feeling for what it actually is.

So, question here is,

What are Lie groupoids? How similar are they to Lie groups, Groupoids and what can one expect to do on a Lie groupoid?

Any reference is appreciated.

5 Answers

Here is an expansion of my comment, by request. The 2-category of Lie groupoids $mathrm{LieGpd}$ admits the category of manifolds $mathrm{Mfld}$ as a full sub-2-category (i.e. $mathrm{Mfld} to mathrm{LieGpd}$ induces an iso on hom-groupoids), but the category of Lie groups $mathrm{LieGrp}$ only has a functor $mathrm{LieGrp} to mathrm{LieGpd}$, not full on 2-arrows, sending a group to the coresponding one-object groupoid. I think this is one reason why Lie groupoids are not best seen as a generalisation of Lie groups.

Given a group $G$ acting on $M$, there is a groupoid $M//G$ with objects $M$, morphisms $Mtimes G$ with source and target given by projection and action, respectively. This includes the boring case where $M=*$, but the mix of potentially positive dimensional orbit space $M/G$, interesting orbits $mG$ nontrivial stabilisers $G_m$ is more typical of what is happening in the Lie groupoid case. Nice Lie groupoids, for instance proper Lie groupoids $X$, where $(s,t)colon X_1to X_0times X_0$ is a proper map, locally look like group actions (times a codiscrete groupoid $Utimes U rightrightarrows U$), so this is sometimes a lot of what you need to consider anyway. Functors are just equivariant maps, where you can be equivariant with respect to a homomorphism $Gto H$.

Generalising from this to general Lie groupoids you need to forget the fact that the source map has identical fibres, and that there isn't some global group containing all the stabilisers.

Correct answer by David Roberts on September 30, 2020

Here is a somewhat trivial observation, but one that helped me to get an initial 'visual' feeling of the thing. I tried to make it a comment, but since I have just signed up, I didn't have enough reputation for doing so. If someone consider it more appropriate to move it somewhere else, please do it. =]

When dealing with purely algebraic groupoids, it is common to call'em 'connected' when two objects are connected by at least one morphism. Also, it is usually OK to deal only with connected groupoids, since the disconnected ones are merely, well, 'disjoint unions'; unless, of course, the groupoid in question had emerged from some problem for which the existence of connected components, and the variability of its isotropy groups, might be relevant.

In the case of Lie groupoids, however, this concept of connectedness doesn't make much sense. Indeed, the strictly analogous condition, i.e., each pair of objects being connected by at least one morphism, is in that case called 'transitiveness' of a groupoid (see, for instance, Mackenzie's book 'General theory of Lie groupoids and Lie algebroids'). 'Connectedness', therefore, is in the context of Lie groupoids thought in relation to the topological structure of the groupoid, which is absent in the purely algebraic context.

Thus, it seems that an important part of the richness of the Lie groupoid theory resides in the fact that many Lie groupoids are not transitive, but still are connected. This means: the 'connected components', in the purely algebraic sense, of a Lie groupoid are somehow 'glued together' in the topological structure of the Lie groupoid.

Answered by Dry Bones on September 30, 2020

The question in the title is not well-defined in my view. Concerning the other questions:

What are Lie groupoids?

Answer: Groupoids internal to the / a category of smooth manifolds, where the source map (equivalently the range map) is assumed to be an admissible submersion. Here by admissible submersion one means submersions which make the pullback structure of the composable arrows of the groupoid again a smooth manifold. Also, I wrote the or a category of smooth manifolds, because what specific category of smooth manifolds you consider is up to you, but then the kind of admissible submersions also varies, depending on the kind of smooth manifolds one considers. Consider for example a category of compact manifolds with boundary (or corners) where the smooth structure is induced by pulling back the smooth structure of a surrounding neighborhood (without corners). Then one needs to be careful to restrict the class of allowed submersions, because otherwise the pullbacks won't exist anymore within the category.

How similar are they to Lie groups?

Answer: As already mentioned in a previous answer a good way to think about groupoids is that they are structures which interpolate between sets and groups. By extension Lie groupoids would be structures which interpolate between smooth manifolds and Lie groups. Perhaps that is the intuition you are looking for? On the other hand, consider a Lie groupoid with set of objects consisting of a single point, this gives you a Lie group (EDIT: up to a strict isomorphism of Lie groupoids). The Lie algebroid of this is (up to isomorphism) the Lie algebra of the Lie group. One well-known important difference to the theory of Lie groups is the integration problem: Given a Lie algebroid $mathcal{A}$ on $M$ it is not always possible to find a Lie groupoid $mathcal{G} rightrightarrows M$, such that $mathcal{A}(mathcal{G}) cong mathcal{A}$.

What can one expect to do on a Lie groupoid?

Answer: Geometry, Analysis and Index theory come to my mind. In Analysis Lie groupoids are used to study partial differential equations on foliated manifolds. Likewise, in geometry, a Lie groupoid can be viewed as a "desingularization" of various types of foliations: Here the leaves of the foliation correspond to the orbits of an underlying Lie groupoid. This is the case whenever the foliation possesses a holonomy groupoid. In the other direction: the orbits of a Lie groupoid always form a singular Stefan-Sussmann foliation. In index theory Lie groupoids are used to obtain proofs of the Atiyah-Singer index theorem and to obtain considerable generalizations of the Atiyah-Singer index theorem (and Poincare duality for K-homology theories on Lie groupoids). Related to this is the role of Lie groupoids in the problem of taking colimits within the category of smooth manifolds, in particular quotients by equivalence relations. Since such quotients usually don't stay within the category of smooth manifolds, a nice idea has been to consider instead the equivalence relation itself as a groupoid. I recommend as reference: Connes, Noncommutative Geometry, Chapter II.

Answered by santker heboln on September 30, 2020

In the same way as groupoids $(mathrm{src},mathrm{trg}):X_1rightrightarrows X_0$ are a symultaneous generalization of group actions

$$(mathrm{pr}_1,mathrm{act}):Gtimes Xrightrightarrows X$$

and equivalence relations

$$(mathrm{pr}_1|_R,mathrm{pr}_2|_R):Rrightrightarrows X;,qquad Rsubseteq Xtimes X$$

on sets $X$, Lie groupoids are a simultaneous generalization of (smooth) Lie group actions and "smooth" equivalence relations (where "smooth" means the maps induced on $Rsubseteq Mtimes M$ by the projections to the first and second factor are submersions) on manifolds $M$.

Group actions and equivalence relations, hence also groupoids, should have quotients, but a quotient manifold does not always literally exist; a Morita morphism between two groupoids is intuitively a smooth morphism between the putative quotient manifolds. The geometric objects that replace the non existing quotient manifolds are called (differentiable) stacks.

Answered by Qfwfq on September 30, 2020

Perhaps you should look through some of the papers on Ronnie Brown's website. In particular

Lie groupoids came from Ehresmann's work on fibre bundles in diff. geom. and were taken forward by Pradines and his ideas on Monodromy and Holonomy groupoids (The exact references to those are in Ronnie Brown's article, and also in the reference below.)

Also of use would be Kirill Mackenzie's introduction

and various parts of that book may help.

I hope this helps.

Answered by Tim Porter on September 30, 2020

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