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I am familiar with the notion of Lie groupoids.

But, only easy examples of Lie groupoids I am familiar with are the following:

  1. Lie groupoids coming from manifolds; that are of the form $(M\rightrightarrows M)$.
  2. Lie groupoids coming from groups; that are of the form $(G\rightrightarrows *)$.
  3. Lie groupoids coming from an action of Lie group on a manifold, say $M\times G\rightarrow M$; that are of the form $(M\times G\rightrightarrows M)$, also called as translation Lie groupoid.

To understand some structure over a Lie groupoid, I would first see their special cases in above examples. This gives some understanding of what the structure is in some special cases. There is a theorem by Ieke Moerdijk and D. A. Pronk that says that any proper étale Lie groupoid is locally a translation groupoid.

Are there other simpler Lie groupoids that you use in the same line of above mentioned Lie groupoids that gives a better understanding of a setup on an arbitrary Lie groupoid?

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  • $\begingroup$ I am not 100% sure that this is well posed question. Please ask if I should add any extra information $\endgroup$ – Praphulla Koushik Nov 15 '19 at 13:41
  • $\begingroup$ Please do not read this question as "what are the examples of Lie groupoids?".. This is not just about examples.. $\endgroup$ – Praphulla Koushik Nov 15 '19 at 14:39
  • $\begingroup$ Not just proper étale Lie groupoids, but arbitrary Lie groupoids look locally like the action groupoid of a compact Lie group $G$ acting on a ball, with this action arising from an orthogonal representation of $G$. $\endgroup$ – David Roberts Nov 20 '19 at 8:11
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    $\begingroup$ @DavidRoberts please give reference for that general result, I only know Moerdijk and Pronk's paper where they say for proper etale Lie groupoids... Do you have any comments on what examples should one consider in checking some results on arbitrary Lie groupoids? $\endgroup$ – Praphulla Koushik Nov 20 '19 at 9:10
  • $\begingroup$ See arxiv.org/abs/1101.0180 and arxiv.org/abs/1103.5245 for the references. Sorry, nothing on examples for now. $\endgroup$ – David Roberts Nov 20 '19 at 10:25
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Equivalence relations $R\rightrightarrows M$ where $R\subseteq M\times M$, $s=p_1$, $t=p_2$, are groupoids (in sets). To get Lie you need the projections to be manifold submersions. It happens for example when the equivalence relation comes from a (nonsingular) foliation on a manifold.

Edit: I see on wikipedia there's also what they call the "pair groupoid", which is an example of equivalence relation not arising from a (positive dimensional) foliation. $p_1,p_2:M\times M\rightrightarrows M$.

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  • $\begingroup$ Yes, this is would be one simple (not for me) case of Lie groupoids, the Foliation groupoid... Do you have any favourite paper or book on Foliations which are not standard references? I know about Moerdijk's book and also there is some chapter in Lee's smooth manifolds.. $\endgroup$ – Praphulla Koushik Nov 15 '19 at 14:36
  • $\begingroup$ It's probably standard for those who work on foliations, but you can have a look at bookstore.ams.org/gsm-23 $\endgroup$ – Qfwfq Nov 15 '19 at 14:38
  • $\begingroup$ I. do not see any relationship between pair groupoid and the trivial groupid $(G\rightrightarrows *)$.. Is it straight forward? $\endgroup$ – Praphulla Koushik Nov 15 '19 at 14:40
  • $\begingroup$ Ah, in the context of foliations you have also the "holonomy groupoid": if I remember well, objects are germs of transversals and isomorphisms are germs of local diffeomorphisms of transversals induced by the holonomy along the leaves. But I wouldn't bet it's a finite dimensional / second countable Lie groupoid at all. $\endgroup$ – Qfwfq Nov 15 '19 at 14:43
  • $\begingroup$ Yes, that's another example... Holonomy groupoid... Does people study some geometric objects w.r.to Holonony groupoids? Do you know some? Thanks for the book link. I will see. $\endgroup$ – Praphulla Koushik Nov 15 '19 at 14:47
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You can restrict your example 3 to a submanifold $N\subseteq M$. The groupoid consists of $(g,m)\in G\times N$ such that $g\cdot m \in N$.

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  • $\begingroup$ I am having difficulty in understanding how this is different from example $3$? :O $\endgroup$ – Praphulla Koushik Nov 25 '19 at 11:50
  • $\begingroup$ What do you mean? If $N$ is not closed under $G$-action, it is not an action groupoid of $G$. In general, it could be quite complicated groupoid. $\endgroup$ – Bugs Bunny Nov 25 '19 at 11:58
  • $\begingroup$ you are mentioning the Lie groupoid $[G\times N\rightrightarrows N]$.. Isn't it? I do not understand the point of considering a submanifold that is invariant under action of G and then considering the associated action Lie groupoid.. This is already covered in example 3. Am I missing something? $\endgroup$ – Praphulla Koushik Nov 25 '19 at 11:59
  • $\begingroup$ No, I am not. In general, $G$ does not act on $N$, only on $M$. So there is no such groupoid. $\endgroup$ – Bugs Bunny Nov 25 '19 at 12:02
  • $\begingroup$ I am really missing something. I will take a break and respond to your answer... You are writing $(g,m)\in G\times N$ with $g.m\in N$... I am seeing this as action of $G$ on $N$... May be I need some fresh air... $\endgroup$ – Praphulla Koushik Nov 25 '19 at 12:12
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My answer somewhat expands on Qfwfq's above. Lie groupoids are a useful tool to reduce certain infinite-dimensional transformation groups to a finite-dimensional setting. Moreover, as put by Alan Weinstein, they are a means of unifying the description of "internal" and "external" symmetries. As a typical example, which happens to be of importance in physics, one may consider $G$-bundle automorphisms $\psi$ of a principal $G$-bundle $\pi:P\rightarrow M$ over $M$, with the structure group $G$ being a finite-dimensional Lie group (from now on, we denote the projections of all $G$-bundles associated to $\pi$ collectively by $\tilde{\pi}$). This means that $\psi$ is a diffeomorphism of $P$ which commutes with the right action of $G$ on $P$. Since the orbits of this action are precisely the fibers of $P$, this entails that $\psi$ maps fibers of $\pi$ onto fibers of $\pi$ and therefore there is a unique diffeomorphism $\psi_M$ of $M$ such that $\pi\circ\psi=\psi_M\circ\pi$ (hence $\psi$ is a bundle automorphism of $P$ in the broader sense, covering $\psi_M$). If $\psi_M=\mathrm{id}_M$, we say that $\psi$ is a strict $G$-bundle automorphism of $\pi$ or a gauge transformation of $\pi$. Therefore, one can define the following three infinite-dimensional transformation groups:

  • $\mathrm{Aut}_G(\pi)=$ group of $G$-bundle automorphisms of $\pi$;

  • $\mathrm{Aut}^0_G(\pi)=$ group of gauge transformations of $\pi$;

  • $\mathrm{Diff}(M)=$ group of diffeomorphisms of $M$, which appear naturally when considering any Lie groupoid over $M$ since by definition (see below) smooth bisections of Lie groupoids over $M$ yield elements of $\mathrm{Diff}(M)$ after composition with the latter's target maps. Particularly, $\mathrm{Diff}(M)$ is (up to a group isomorphism) the space of smooth bisections of the pair groupoid of $M$.

The first two act not only on $\pi$ but also on any $G$-bundle associated to $\pi$. It turns out that $\mathrm{Aut}^0_G(\pi)$ is a normal subgroup of $\mathrm{Aut}_G(\pi)$ and $\mathrm{Aut}_G(\pi)/\mathrm{Aut}^0_G(\pi)\cong\mathrm{Diff}(M)$ (the isomorphism being the map $[\psi]\rightarrow\psi_M$, where $[\psi]$ is the equivalence class of $\psi\in\mathrm{Aut}_G(\pi)$ modulo $\mathrm{Aut}^0_G(\pi)$), but $\mathrm{Diff}(M)$ is not generally (isomorphic to) a normal subgroup of $\mathrm{Aut}_G(\pi)$. A(n essentially) finite-dimensional treatment of $\mathrm{Aut}_G(\pi)$ and $\mathrm{Aut}^0_G(\pi)$ is made possible through the following finite-dimensional Lie groupoids, similarly as we expressed $\mathrm{Diff}(M)$ above as space of smooth bisections of the pair groupoid $M\times M\rightrightarrows M$. The idea is that the action of the above infinite-dimensional transformation groups can be reduced for most purposes to the action of the corresponding (finite-dimensional) Lie groupoids.

  • Take the orbit space $(P\times P)/G$ of $P\times P$ modulo the diagonal action of $G$ on $P\times P$ induced by the right action of $G$ on $P$: $$(q_1,q_2)\cong(q_1,q_2)g\doteq(q_1g,q_2g)\ ,\quad g\in G\ ,\,q_1,q_2\in P\ .$$ Since $\pi\circ\mathrm{pr}_j(q_1,q_2)=\pi\circ\mathrm{pr}_j(q_1g,q_2g)=\pi(q_j)$ for all $q_1,q_2\in P$, $g\in G$, $j=1,2$, we see that $\pi\circ\mathrm{pr}_1,\pi\circ\mathrm{pr}_2:P\times P\rightarrow M$ respectively induce unique source and target maps $\sigma,\tau:(P\times P)/G\rightarrow M$ given by $\sigma((q_1,q_2)G)=\pi(q_1)$, $\tau((q_1,q_2)G)=\pi(q_2)$. Moreover, the pair groupoid multiplication $(q_1,q_2)\cdot(q_2,q_3)=(q_1,q_3)$ on $P\times P$ induces a multiplication map $$\mu:(P\times P)/G\,{}_\tau\!\!\times_\sigma(P\times P)/G\doteq\{((q_1,q_2)G,(q_3,q_4)G)\in(P\times P)/G\ |\ \pi(q_2)=\pi(q_3)\}\rightarrow(P\times P)/G$$ given by $$\mu((q_1,q_2)G,(q_3,q_4)G)=(q_1,q_4)G$$ whenever $\tau((q_1,q_2)G)=\pi(q_2)=\sigma((q_3,q_4)G)=\pi(q_3)$. Finally the composition of the diagonal map from $P$ to $P\times P$ with the quotient map modulo the diagonal action of $G$ on $P\times P$ defines a unit map $$\iota:M\ni p\mapsto(q,q)G\in(P\times P)/G$$ (where $q$ is any fixed element of $\pi^{-1}(p)$) and the inversion map $$\nu:(P\times P)/G\ni(q_1,q_2)G\mapsto (q_2,q_1)G\in(P\times P)/G$$ is then given in the obvious way. These five maps together define a Lie groupoid structure on $(P\times P)/G$, which is then called the gauge groupoid $\mathrm{Gau}_G(\pi)$ of $\pi$.

  • The so-called strict or internal gauge groupoid $\mathrm{Gau}^0_G(\pi)$ of $\pi$ is simply the (associated $G$-)bundle of Lie groups $\tilde{\pi}:P\times_G G\rightarrow M$, with $G$ acting on itself by conjugation (this is sometimes called the adjoint bundle of $\pi$). Recall that $P\times_G G$ is the orbit space of $P\times G$ under the (right) $G$-action $$(q,h)g=(qg,g^{-1}hg)\ ,\quad q\in P\ ,\,g,h\in G\ .$$ This bundle embeds naturally into $(P\times P)/G$ through the map $$\epsilon:P\times G\ni(q,h)\mapsto\epsilon(q,h)\doteq(q,qh)\in P\times P\ ,$$ which is clearly seen to be compatible with the respective $G$-actions: $$\epsilon((q,h)g)=\epsilon(qg,g^{-1}hg)=(qg,qgg^{-1}hg)=(q,qh)g=\epsilon(q,h)g\ .$$ The image of the induced embedding $$\tilde{\epsilon}:P\times_G G\ni(q,h)G\mapsto\tilde{\epsilon}((q,h)G)\doteq(q,qh)G\ni(P\times P)/G$$ is simply the subset of $(P\times P)/G$ to which the restrictions of $\sigma$ and $\tau$ coincide - that is, the isotropy subgroupoid of $\mathrm{Gau}_G(\pi)$. As such, we can set the source and target maps $\sigma,\tau$ of $\mathrm{Gau}^0_G(\pi)$ as $$\sigma((q,h)G)=\tau((q,h)G)\doteq\pi(q)$$ and the multiplication map $\mu$ in $P\times_G G$ as $\mu((q_1,h_1)G,(q_2,h_2)G)\doteq(q_1,gh_1h_2)G$ whenever $\tilde{\pi}((q_1,h_1)G)=\pi(q_1)=\tilde{\pi}((q_2,h_2)G)=\pi(q_2)$, which amounts to $q_2=q_1g$ for a unique $g\in G$. Moreover, we can also set the unit map $$\iota:M\ni p\mapsto\iota(p)=(q,e)G\in P\times_G G\ ,$$ where $q$ is any element of $\pi^{-1}(p)$ and $e$ is the unit of $G$, and the inversion map $$\nu:P\times_G G\ni(q,h)G\mapsto\nu((q,h)G)=(q,h^{-1})G=(qh,e)G\in P\times_G G\ .$$ If we set $\mathrm{Gau}^0_G(\pi)$ as $P\times_G G$ endowed with such Lie groupoid operations, $\tilde{\epsilon}$ as defined above becomes a Lie groupoid (mono)morphism. More generally, a bundle of Lie groups over $M$ is always a Lie groupoid over $M$ for which the source and target maps coincide (with the bundle projection being, of course, any of those two), that is, a Lie groupoid which coincides with its own isotropy subgroupoid.

To make the connection of $\mathrm{Gau}_G(\pi)$ to $\mathrm{Aut}_G(\pi)$ and of $\mathrm{Gau}^0_G(\pi)$ to $\mathrm{Aut}^0_G(\pi)$, recall that a smooth bisection $\gamma:M\rightarrow G_1$ of a Lie groupoid $(\sigma,\tau):G_1\rightrightarrows M$ over $M$ is a smooth map such that $\sigma\circ\gamma=\mathrm{id}_M$ and $\tau\circ\gamma\in\mathrm{Diff}(M)$. If we denote by $\Gamma(G_1\rightrightarrows M)$ the space of smooth bisections of $G_1\rightrightarrows M$, we see that such a space is always a group if we set their product as follows: if $\gamma_1,\gamma_2\in\Gamma(G_1\rightrightarrows M)$, we write $(\gamma_1\gamma_2)(p)=\gamma_1\circ\tau\circ\gamma_2(p)$. The identity of $\Gamma(G_1\rightrightarrows M)$ is, of course, the unit map $\iota$, and the inverse of $\gamma\in\Gamma(G_1\rightrightarrows M)$ is given by $\gamma^{-1}=\nu\circ\gamma$, where $\nu$ is the inversion map. We can then establish the following group isomorphisms:

  • $\mathrm{Aut}_G(\pi)\cong\Gamma(\mathrm{Gau}_G(\pi)\rightrightarrows M)$ through $\psi\mapsto(p\mapsto(q,\psi(q))G)$, where $q$ is any element of $\pi^{-1}(p)$. This yields $\sigma((q,\psi(q))G)=\pi(q)=p$ and $\tau((q,\psi(q))G)=\pi\circ\psi(q)=\psi_M\circ\pi(q)=\psi_M(p)$, as desired.

  • $\mathrm{Aut}^0_G(\pi)\cong\Gamma(\mathrm{Gau}^0_G(\pi)\rightrightarrows M)$ through $\psi\mapsto(p\mapsto(q,g_q)G)$, where $q$ is any element of $\pi^{-1}(p)$ and $g_q\in G$ is the unique element such that $\psi(q)=qg_q$. Thus, $\sigma((q,g_q)G)=\tau((q,g_q)G)=\pi(q)=p$, as desired. We notice as well that $\Gamma(\mathrm{Gau}^0_G(\pi)\rightrightarrows M)=\Gamma(P\times_G G\rightarrow M)$.

  • $\mathrm{Diff}(M)\cong\Gamma(M\times M\rightrightarrows M)$. We have already alluded to this isomorphism above, which is given by $\psi_M\mapsto(p\mapsto(p,\psi_M(p)))$.

In other words, we are in a sense "factoring out" the action of the structure group $G$ on the pair groupoid of $P$. This may restrict the elements of $\mathrm{Diff}(M)$ which appear after composition of smooth bisections with $\tau$ - for instance, if we set $\pi$ as the orthonormal frame bundle of the Riemannian manifold $(M,g)$ (where $g$ is some fixed Riemannian metric on $M$), the only diffeomorphisms of $M$ which appear in this way are the isometries of $(M,g)$.

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  • $\begingroup$ Thank you so much for your answer. I am sure you spent so much time in writing this down... Please give me some time to respond.. $\endgroup$ – Praphulla Koushik Nov 27 '19 at 6:37

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