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Let $A\rightrightarrows X$ be a groupoid, where $X$ is the set of objects and $A$ is the set of arrows. My favorite example of a groupoid is an action groupoid. If a group $G$ acts on the left on a set $X$, we set $$ A=\{(x,g,y)\mid x,y\in X, g\in G, y=g*x\}, $$ then $A\rightrightarrows X$ with the evident maps is called the action groupoid corresponding to the action of $G$ on $X$. It is often denoted by $G\ltimes X$. If $G$ acts on $X$ transitively, then $G\ltimes X$ is a connected groupoid. Conversely, any connected groupoid is isomorphic to an action groupoid, see the answers to my question Connected groupoids and action groupoids.

Now let $\Gamma$ be a group, and assume that $\Gamma$ acts compatibly on $G$ and on $X$ (see my question Equivalence and weak equivalence of groupoids for a natural example of such action). We say that the $\Gamma$-groupoid $G\ltimes X$ is an action $\Gamma$-groupoid.

Question 1. Is it true that any connected $\Gamma$-groupoid is isomorphic to an action $\Gamma$-groupoid?

Question 2. Is it true that any connected $\Gamma$-groupoid is weakly $\Gamma$-equivalent to an action $\Gamma$-groupoid?

See Equivalence and weak equivalence of groupoids for the definition of a quasi-isomorphism (weak equivalence). We say that two $\Gamma$-groupoids are weakly $\Gamma$-equivalent if they can be connected by a chain of quasi-isomorphisms of $\Gamma$-groupoids.

I expect the answer "No" to Question 1, and therefore I ask Question 2, to which I expect the answer "Yes".

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2 Answers 2

The answer to Question 1 is, as expected, "No".

Let $(A\rightrightarrows X)=(X\times X\rightrightarrows X)$. If our groupoid is an action groupoid, i.e., of the form $G\ltimes X$, then the $\Gamma$-set $X$ with $G$-action must be a torsor (principal homogeneous space) of $G$.

Take $X=\{1,2,\dots,p\}$, where $p$ is a prime number, $p\ge 5$. Take $\Gamma=A_p$ (the alternating group) with the standard action on $X$. We show that our $\Gamma$-set $X$ cannot be isomorphic to a torsor of a $\Gamma$-group.

If our $X$ is a torsor of a $\Gamma$-group $G$, then $|G|=p$, hence $G\simeq{{\mathbb{Z}}}/p{{\mathbb{Z}}}$ and ${\rm Aut}\ G\simeq {{\mathbb{F}}}_p^\times$. Since $\Gamma$ is a simple group, there are no non-trivial homomorphisms $\Gamma\to {{\mathbb{F}}}_p^\times$, hence $\Gamma$ acts on $G$ trivially. Torsors of $G$ are classified by $$ H^1(\Gamma, G)={\rm Hom}(\Gamma, {{\mathbb{Z}}}/p{{\mathbb{Z}}})=0. $$ We conclude that if $|X|=p$ and $X$ is a torsor of a $\Gamma$-group with $\Gamma=A_p$, then $\Gamma$ acts on $X$ trivially. Thus our groupoid $X\times X\rightrightarrows X$, where $\Gamma$ acts on $X$ in the standard way (hence, nontrivially), is not isomorphic to an action groupoid.

The above argument goes through for $X=\{1,2,\dots,n\}$ and $\Gamma=A_n$ for any $n\ge5$, not necessarily prime. Indeed, then ${\rm Aut}\ G\subset S_{n-1}$, hence there are no nontrivial homomorphisms $\Gamma\to{\rm Aut}\ G$.

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I might be wrong but at first glance, I would say "yes" to Question 1. The reason is the following. If a groupoid $\Gamma \rightrightarrows Y$ acts by automorphisms on a groupoid $A\rightrightarrows X$, one can form the semi-direct product groupoid $\Gamma\ltimes A$ with unit space $X$ and morphisms $(\Gamma\ltimes A)_1$ elements $(\gamma,a)\in \Gamma\times A$ such that $\gamma a$ makes sense. Now, if $A\cong G\ltimes X$ is connected, and if $\Gamma$ is a group acting compatibly on $G$ and $X$, then $\Gamma\ltimes (G\ltimes X)\rightrightarrows X$ is isomorphic to $(\Gamma\ltimes G)\ltimes X \rightrightarrows X$. Moreover, any connected $\Gamma$-groupoid is in particular connected and hence in the form $G\ltimes X$. By construction of the group $G$, $\Gamma$ acts compatibly on it and $X$, and $\Gamma\ltimes A$ is a action $\Gamma$-groupoid.

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@Mkouboi: I do not understand your answer. You write: "Moreover, any connected Γ-groupoid is in particular connected and hence in the form G⋉X. By construction of the group G, Γ acts compatibly on it and X..." But the construction of $G$ from the groupoid A⇉X is not functorial, so I do not understand how you get an action of Γ on G. Could you please add details? –  Mikhail Borovoi May 17 '13 at 17:07
    
@Mikhail: A groupoid isomorphism from $A\rightrightarrows X$ to $G\ltimes X\rightrightarrows X$ induces a $\Gamma$-action on the group bundle $G\ltimes X$. One obtains the $\Gamma$-action on $G$ by identifying $G$ with the fibres $(G\ltimes X)_x=(G\ltimes X)^x=(G\ltimes X)^x_x$. –  Mkouboi May 18 '13 at 16:08
    
@Mkouboi: I do not assume that there is a $\Gamma$-fixed point in $X$, and therefore $\Gamma$ does not act on the fibres $A(x,-)$ or $A(-,x)$ or $A(x,x)$. –  Mikhail Borovoi May 18 '13 at 16:45
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