# Condition on a Lie groupoid to be represented by manifold/group or an action groupoid

Let $$\mathcal{G}$$ be a Lie groupoid. I am thinking of following questions.

1. When do we know $$\mathcal{G}$$ is weakly/Morita equivalent to a Lie groupoid of the form $$(G\rightrightarrows *)$$ for some Lie group $$G$$?
2. When do we know $$\mathcal{G}$$ is weakly/Morita equivalent to a Lie groupoid of the form $$(M\rightrightarrows M)$$ for some manifold $$M$$?
3. When do we know $$\mathcal{G}$$ is weakly/Morita equivalent to a Lie groupoid of the form $$G\ltimes M$$ for some Lie group $$G$$ acting on a manifold $$M$$?

It is only reasonable to see what properties the Lie groupoids $$(G\rightrightarrows *), (M\rightrightarrows ), G\ltimes M$$ have so that we can say $$\mathcal{G}$$ should have atleast these properties.

My observations:

1. Given a Lie group $$G$$, the Lie groupoid $$(G\rightrightarrows *)$$ is transitive Lie groupoid. Thus, it should be the case that $$\mathcal{G}$$ is necessarily transitive to be represented by Lie group. Surprisingly, any transitive Lie groupoid is representable by a Lie group, more precisely $$\mathcal{G}\cong \mathcal{G}_x$$ (isotropy group) for any $$x\in \mathcal{G}_0$$ . I do not know if we can say anything more in this. Is there any other characterization? Can we say anything more?
2. For a manifold $$M$$, the Lie groupoid $$(M\rightrightarrows M)$$ is a proper Lie groupoid simply because, $$(s,t):M\rightarrow M\times M$$ is the diagonal map which is a proper map. So, $$(M\rightrightarrows M)$$ is a proper Lie groupoid. If we want $$\mathcal{G}$$ to be represented by manifold, we should have atleast this condition that $$\mathcal{G}=(P\rightrightarrows X)$$ is a proper Lie groupoid i.e., $$(s,t):P\rightarrow X\times X$$ is a proper map. For manifold $$M$$, $$(s,t):M\rightarrow M\times M$$ is injective. If we want $$\mathcal{G}$$ to be represented by manifold, we should have atleast this condition that $$\mathcal{G}=(P\rightrightarrows X)$$ is such that $$(s,t):P\rightarrow X\times X$$ is injective. David Roberts says here that, assuming that $$(s,t):P\rightarrow X\times X$$ is proper and injective sufficient to confirm $$\mathcal{G}\cong X/P$$ (Any quick proof for this is also welcome). Are there any other characterization? How do we know where to stop looking at properties for characterization? I mean how do we guess (before proving) that $$(s,t)$$ is proper and injective then, $$\mathcal{G}$$ is represented by manifold. Why proper, injective determining "manifoldness" of Lie groupoid?

Are there similar characterizations for a Lie groupoid to be a translation groupoid i.e., of the form $$G\ltimes M$$ for some Lie group $$G$$ acting on manifold $$M$$? It is observed here that, if a Lie groupoid is proper and etale, then $$\mathcal{G}$$ is locally a translation groupoid. Are there any characterization for whole Lie groupoid to be translation groupoid? What should we look for in a Lie groupoid to expect it to be a translation groupoid?

• I have conjectured a long time ago that every compact orbifold is (Morita equivalent to) a quotient of a manifold by a compact Lie group. See my PhD thesis available at andreghenriques.com/thesiswarning.html. I strongly believe that this conjecture is true. But my understanding is that this conjecture is still open. Note that, for effective orbifolds, the statement is very easy to prove as the total space of the frame bundle is then a manifold. – André Henriques Feb 16 at 23:41
• @André I guess by orbifold you mean a proper étale Lie groupoid, as opposed to the traditional definition? – David Roberts Feb 17 at 1:06
• @PraphullaKoushik Did you read the warning he wrote on the page he linked? – S. Carnahan Feb 17 at 6:08
• @PraphullaKoushik I have deleted your request for current status, because it is already answered in André's comment. – S. Carnahan Feb 17 at 13:38
• @Prapulla Koushik. Yes, despite what is written in that section, the conjecture is still open. I have spent a large amount of time and energy in the years following my PhD trying to complete the proof. But I have not succeeded. It's now time for others to try to do what I have failed to achieve. (And I'm of course happy to offer advice to whoever wishes to try) – André Henriques Feb 17 at 18:20