I am trying to understand the semidirect product of groupoids, as defined in this answer by Theo Johnson-Freyd. Part of my difficulty is that although the definition of a 2-group makes sense, I am quite unfamiliar with that concept. So, as a test, I 'computed' the crossed product of two (arbitrary) sets $X,Y$ seen as discrete groupoids. My understanding is that the 2-group of automorphism in this case is the trivial monoidal category $(\mathbf{1}, \otimes, \ast)$, with $\mathbf{1}$ the trivial 1-object category, $\otimes:\mathbf{1}\times\mathbf{1}\rightarrow\mathbf{1}$, and unit the unique object $\ast$.

The object set of $X \rtimes Y$ is then $X \times Y$, and there is an arrow $(x_1,y_1) \rightarrow (x_2,y_2)$ iff $x_1=y_1$ and $x_2=y_2$, a long winded way of saying that in this particular case, $X \rtimes Y$ is (exactly) $X\times Y$ as a discrete groupoid. Question 1: is this correct?

There is another case that I have not been able to work as successfully: given an equivalence relation $E_1$ on (set) $X$ seen as a groupoid, and equivalence relation $E_2$ on $Y$, what are $E_1 \rtimes E_2$ and $E_2 \rtimes E_1$ ?

[Edit: I meant semidirect product, not crossed]


An action of a groupoid on another groupoid was defined in my paper "Groupoids as coefficients" Proc LMS (3) 25 (1972) 413-426, available here, which also uses methods of fibrations of groupoids. The term "coefficients" refers to nonabelian cohomology. This paper uses the term "split extension" instead of semidirect product. The more modern term is used given an action of a group on a groupoid in the book Topology and Groupoids, where it is relevant to the explicit description of orbit groupoids. Note that if a group acts on a space $X$, then it has an induced action on the fundamental groupoid $\pi_1 X$, and also on the groupoid $\pi_1(X,A)$ for any set $A$ of base points which is a union of orbits.

  • $\begingroup$ @Ronnie: Cool, I had been wondering when it was originally defined. I "rediscovered it" (in the topological context) a couple of years ago, but was certain I was not the first to think of it. I was pointed to some other references, but I wasn't sure from where it originally came. $\endgroup$ – David Carchedi Feb 15 '13 at 13:10
  • $\begingroup$ @David: I have not checked this, but Andree Ehresmann referred me to Charles Ehresmann's "Categories et Structure" pp. 49-50. At the time of my paper, I was unaware of Ehresmann's extensive work on groupoids, particularly the topological and differentiable case. What was important for me in his book was the notion of double category, and his example of the commuting squares in a category. $\endgroup$ – Ronnie Brown Feb 15 '13 at 15:48

The "crossed product" is not an operation on groupoids. Given one groupoid $G$, one can consider the category $G-Set$ of sets with a $G$-action (the classifying topos of $G$). Let $H$ be an internal groupoid in the category $G-Set$ (equivalently, a groupoid $H$ with an action of $G$). Then one can form a groupoid $G \ltimes H$ which comes equipped with a canonical functor $G\ltimes H \to G.$ This is actually a construction you (probably) already know:

The category $G-Set$ is the same as the category of functors from $G$ into $Set,$ or from $G^{op}$ into $Set$ since $G$ is a groupoid, these are the same. Hence, we can identify $G-Set$ with the category of presheaves on $G$. Therefore, $H$ is a groupoid object in presheaves, which is the same as a presheaf of groupoids. The functor $$G\ltimes H \to G$$ is the fibration arising from applying the Grothendieck construction of this presheaf of groupoids.

I give a description of this construction in the case of topological groupoids (under the name generalized action groupoids) here:


(See page 24)

  • $\begingroup$ I don't understand your answer at all, at least in the context of my question [which reflects on me, not your answer]. I took the definition from the answer I linked to, and carried out the computation, everything 'worked'. I just realized one mistake: I said 'crossed' when I meant 'semidirect'. But I get the feeling that that was not crucial. $\endgroup$ – Jacques Carette Feb 14 '13 at 22:23
  • $\begingroup$ Part of my point is that you have a misconception: there is no such thing as the "crossed" or "semidirect" product of two groupoids G and H. In order to form such an operation, you need more data- you also need an action of one of the groupoids on the other one. $\endgroup$ – David Carchedi Feb 14 '13 at 23:05
  • $\begingroup$ Right. This worked out in my overly-simple example because the action was unique (and trivial). Ok, I'll think about that. $\endgroup$ – Jacques Carette Feb 15 '13 at 0:22

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