It is written in Wikipedia http://en.wikipedia.org/wiki/Groupoid, that any connected groupoid $A\rightrightarrows X$ is isomorphic to an action groupoid $G\ltimes X$ coming from a transitive action of some group $G$ on $X$. I do not understand how to construct such a group $G$, and would be grateful for an explanation or a reference.

I think that in general one cannot recover $G$ from the action groupoid $G\ltimes X$. Indeed, if $G$ acts simply transitively on $X$, then the action groupoid is given by the equivalence relation $X\times X$ on $X$, hence does not depend on $G$, provided that ${\rm Card}(G)={\rm Card}(X)$. Is this correct?

This question is a version of my question at Math Stack Exchange to which I got no answers.

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    $\begingroup$ I answered on math.stack exchange. $\endgroup$ – Omar Antolín-Camarena Apr 17 '13 at 4:21
  • $\begingroup$ This is easier than everyone is making it sound. $\endgroup$ – Omar Antolín-Camarena Apr 17 '13 at 4:23

Here's what I wrote on Math Stack Exchange:

A connected groupoid A can be written as an action groupoid for many different groups G. All the groupoid determines is H, the group of automorphisms of any object in the groupoid, and the index of H in G, which is the cardinality of the set of objects of the groupoid. And any group G with subgroup H of the correct index, the action of G on the set of cosets of H has action groupoid isomorphic to A.

This is to be expected, because if we think of H as a one object groupoid and of X as an indiscrete groupoid (the set of objects is X, and there is a unique morphism between any pair of objects) then the original groupoid A is isomorphic to the product H × X, so the isomorphism class of A depends only on the group H and the cardinality of X.

As an extreme example of this, let G act on itself by translation and take the action groupoid. This has set of objects G and a unique morphism between every pair of elements. Notice all traces of the group structure of G are gone: the isomorphism class of this indiscrete groupoid only depends on the cardinality of G.

UPDATE 2: Here is a sloganized answer to the question: the equivalence class of a connected groupoid A is determined by the isomorphism class of the group H = Aut(x0); the isomorphism class of a category is given by the data of its equivalence class plus the number of isomorphic copies of each object in a skeleton.

UPDATE: Here is a proof of the claims above "UPDATE 2".

Claim 1: A is isomorphic to H × X.

Proof. Choose an object x0 of A, identify H with Aut(x0) and choose arbitrary morphisms ax : x0x. The isomorphism H × XA is the identity on objects and sends a morphism (h, u) : xy to the morphism ay h ax-1. (Here u is the unique morphism in X from x to y.) The inverse AH × X sends a morphism a : xy to (ay-1 a ax, u) --same u as above.

Claim 2: For any group G with a subgroup H of index |X|, the action groupoid of G acting on the set G/H of cosets is isomorphic to A.

Proof. Both groupoids are isomorphic to H × X.

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  • $\begingroup$ @Omar: You write: "Any group $G$ with subgroup $H$ of the correct index, the action of $G$ on the set of cosets of $H$ has action groupoid isomorphic to $A$". How can one construct such an isomorphism? $\endgroup$ – Mikhail Borovoi Apr 17 '13 at 8:29
  • $\begingroup$ @Mikhail: Pick a set of representatives for the cosets of H, and pick a bijection between those representatives and the objects of A. I can add more detail later when I'm at a computer with a real keyboard. $\endgroup$ – Omar Antolín-Camarena Apr 17 '13 at 11:20
  • $\begingroup$ @Omar: Thank you, now we have a group $G$ and a action of $G$ on $X$. How can we define an isomorphism between $G\ltimes X$ and $A$? Please kindly add more details when you can. $\endgroup$ – Mikhail Borovoi Apr 17 '13 at 15:19
  • $\begingroup$ I hope that is clear enough, @MikhailBorovoi. If not let me know. As you can see, the proof is pretty much just what Sam Cunningham wrote (only slightly generalized). If you want an explicit isomorphism between A and G/H, you can just compose the isomorphisms A -> H x X -> G/H given in the proof of Claim 1. $\endgroup$ – Omar Antolín-Camarena Apr 18 '13 at 4:41

In the $C^\infty$ case, the Lie groupoid is the action of an identity neighborhood of the Lie group. You can enlarge the action by enlarging $X$, but you may loose Hausdorff. This goes back to Palais. See (and references therein):

  • Franz W. Kamber, Peter W. Michor: Completing Lie algebra actions to Lie group actions. Electron. Res. Announc. Amer. Math. Soc. 10 (2004) 1-10.(pdf)
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Here is my stackexcnage answer.

Another way of looking at this is to use the equivalence of categories between covering morphisms of a groupoid $P$ and actions of $P$ on sets. (Recall that a covering morphism $p:G \to P$ is a groupoid morphism having unique path lifting. Not necessarily unique path lifting gives a fibration of groupoids.) Given an operation of $P$ on a set $X$ then the corresponding covering morphism may be written $P \ltimes X$, an action groupoid, and thought of as a semidirect product because it is a special case of the semidirect product for an action of a groupoid $P$ on a groupoid $H$. For this one needs a morphism of groupoids $\omega: H \to Ob(P)$, where the latter is thought of as a discrete groupoid, and an element $w: x \to y$ in $P$ gives a morphism of groupoids $w_*: \omega^{-1}(x) \to \omega^{-1}(y)$. One has to be precise on conventions to get all this right, which I won't do here.

So a groupoid $G$ has a representation as an action groupoid whenever you are given a covering morphism $ G \to P$. This is closely related to Omar's answer, of course.

I'll add that more details of these ideas are in my book Topology and Groupoids.

Addition: May 19, 2013 Here is a version of Sam's nice argument but in the language of covering morphisms.

If $G$ is a group then its universal cover $p: T \to G$ is a covering morphism of groupoids such that $T$ is connected and has trivial vertex groups; this is determined by the action of $G$ on itself by left multiplication. The set of objects of $T$ is bijective with the set $G$ and $T$ is the indiscrete groupoid (also called tree groupoid) on its set of objects. (Note that if $S$ is a generating set for $G$ then $p^{-1}(S)$ is a graph, namely the Cayley graph of $(G,S)$.)

Now let $A$ be a connected groupoid with $X$ as its set of objects. Let $T$ be the indiscrete groupoid on $X$. Then for any object $x$ of $A$, $A$ is isomorphic to $A(x) \times T$. But if $G$ is as above, then $$1 \times p: A(x) \times T \to A(x) \times G$$ is a covering morphism.

The next question is whether this argument can illuminate the case $A$ is a $\Gamma$-groupoid.

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Given a connected groupoid $A \rightrightarrows X$, pick a basepoint $x_0$ and a morphism $f_{x}: x \to x_0$ for each object $x\in X$. We also need to choose a group structure $(X,.)$ on $X$. Let $G_0 = Aut(x_0)$, and set $G = G_0 \times X$ (as a group).

Then $G$ acts on $X$ by $(g,x) \ast y = x.y$.

Define an isomorphism $\Phi$ from $G \ltimes X$ to $A \rightrightarrows X$ by

$\Phi(y \xrightarrow{(g,x)} x.y)=f_{x.y}^{-1} \circ g \circ f_y$.

This is compatible with composition: If we have morphisms in $G\ltimes X$

$y \xrightarrow{(g,x)} x.y \xrightarrow{(h,w)} w.x.y$,

then applying $\Phi$ gives

$f_{w.x.y}^{-1} \circ h \circ f_{x.y} \circ f_{x.y}^{-1} \circ g \circ f_y = f_{w.x.y}^{-1} \circ hg \circ f_y = \Phi(y \xrightarrow{(hg,x.y)} w.x.y)$

as required.

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  • $\begingroup$ In any case, I think what is written on the wikipedia page is wrong. They seem to want to set $G$ to be $G_0$. But if you have a connected groupoid with no automorphisms, this is not correct (in this case you should just take $G$ to be some group structure on $X$ and use the translation action). $\endgroup$ – Sam Gunningham Apr 16 '13 at 20:11
  • $\begingroup$ These are some of the groups you can use, but in general you don't need $G$ to be a product, just to have a subgroup isomorphic to $G_0$ with index $|X|$. $\endgroup$ – Omar Antolín-Camarena Apr 17 '13 at 4:30
  • $\begingroup$ Sure, I never claimed this was the only way. This was just the first construction that came to mind. $\endgroup$ – Sam Gunningham Apr 17 '13 at 5:49
  • $\begingroup$ But, I agree, your answer is much clearer. $\endgroup$ – Sam Gunningham Apr 17 '13 at 5:50
  • $\begingroup$ @Sam: Thank you for detailed answer! Unfortunately, I cannot accept two answers, otherwise I would accept your answer as well... $\endgroup$ – Mikhail Borovoi Apr 18 '13 at 5:50

The post by John Baez in the $n$-Category Café gives a nice example of how the the Matthieu Group $M_{12}$ can be presented more simply using a groupoid called $M_{13}$.

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