# How can I understand the “groupoid” quotient of a group action as some sort of “product”?

Recall the notion of groupoid (Wikipedia, nLab). An important construction of groupoids is as "action groupoids" for group actions. Namely, let $X$ be a groupoid and $G$ a group, and suppose that $G$ acts on $X$ by groupoid automorphisms. Then we can form a new groupoid $X//G$, which has as objects the objects of $X$, but the morphisms include, in addition to the original morphisms of $X$, a morphism $x \overset g \to gx$ for each $g\in G$ and $x\in X$. The composition of morphisms is well-defined if the action is by groupoid automorphisms. (When $X$ is a set, then $X//G$ is equivalent to the skeletal groupoid whose objects are the elements of the "coarse" quotient $X/G$, and with ${\rm Aut}(\bar x) = {\rm Stab}_G(x)$.)

(Probably there is a fancier construction, in which the conditions on the word "group action" be relaxed to an "action" up to specified natural isomorphism, and then $G$ could act on $X$ by autoequivalences, rather than autoisomorphisms, but this generalization won't concern me.)

Let $1$ denote the one-point set, thought of as a groupoid with only identity morphisms. Then any group $G$ acts uniquely on $1$, and so we have the groupoid $1//G$. In general, although $X\times 1 \cong X$, we do not have $X \times (1//G) \cong X//G$ for arbitrary $G$-actions on $X$ unless the action is trivial. (Here $\times$ denotes the groupoid product, which is just what you think it is.) However, the construction provides natural bijections between the objects of $X//G$ and the objects of $X \times (1//G)$, and between the morphisms of $X//G$ and the morphisms of $X \times (1//G)$.

Question: Is there some sort of "semidirect" or "crossed" product of groupoids, which presumably depends on extra data, so that we do have $X//G \cong X \rtimes (1//G)$? By which I mean, what is the correct notion of "action" of a groupoid $Y$ on a groupoid $X$ and what is the corresponding correct notion of $X \rtimes Y$?

I see that the page semidirect product in nLab defines $X \rtimes G$ as something closely related to $X//G$. But clearly this ought to be called $X\rtimes (1//G)$, but then I do not know what the right definition for $X\rtimes Y$ is, hence the question. And really I'd like to know about a "double crossed product" $X\bowtie Y$.

My motivation for this question is from my answer to Do rational numbers admit a categorification which respects the following “duality”?.

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By the way, Theo, the nLab has a forum where you will be much more likely to get an answer quickly to all of these questions. math.ntnu.no/~stacey/Vanilla/nForum –  Harry Gindi Apr 29 '10 at 21:20
@HG: Thanks! I didn't know about that forum. –  Theo Johnson-Freyd Apr 30 '10 at 3:58
Thanks for asking this question! I was working on it after your answer to my question, but wasn't getting very far. –  Steven Gubkin Apr 30 '10 at 16:02

Let $X$ and $Y$ be groupoids. An action of $Y$ on $X$ is a functor $\rho: Y \to B\operatorname{Aut}(X)$, where $B\operatorname{Aut}(X)$ is the one-object 2-groupoid such that $\operatorname{Hom}(\ast, \ast)$ is the 2-group of autoequivalences of $X$.

We define $X \rtimes Y$ as follows. Its objects are simply $\operatorname{Ob}(X) \times \operatorname{Ob}(Y)$. An element of $(X \rtimes Y)((x_1, y_1), (x_2, y_2))$ consists of a pair $(f, g)$, where $g \in Y(y_1, y_2)$, and $f \in X(x_1, \rho(g)x_2)$. Given $(f, g) \in (X \rtimes Y)((x_1, y_1), (x_2, y_2))$, and $(f', g') \in (X \rtimes Y)((x_2, y_2), (x_3, y_3))$, we define $(f', g') \circ (f, g) \in \operatorname{Hom}((x_1, y_1), (x_3, y_3))$ as $(\rho(g)f' \circ f, g' \circ g)$. It is straightforward to check that in the case that $Y$ is $1 // G$, $X \rtimes Y \cong X // G$.

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Also, this is a special case of the Grothendieck construction (ncatlab.org/nlab/show/Grothendieck%20construction) for pseudofunctors into Cat, where the domain is a groupoid $Y$ and the pseudofunctor happens to take all the objects of $Y$ to the object $X\in Cat$. –  Mike Shulman Apr 30 '10 at 14:29
You can also view the groupoid $X$ together with a $Y$-action as a groupoid in the category $Y-Set \simeq Set^{Y^{op}},$ hence a presheaf of groupoids on $Y,$ and perform the Grothendieck construction to this. –  David Carchedi Feb 14 '13 at 21:05

The notion of semidirect product $\Gamma \rtimes G$ where $G$ is a group acting on a groupoid $\Gamma$ is set up in Chapter11, Section 11.4, of my book "Topology and Groupoids".

It is used there in connection with studying orbit groupoids, and their relevance to the fundamental groupoid of an orbit space by a group action.

One nice point is that this semidirect product includes the case $\Gamma$ is a discrete groupoid, i.e. essentially a set, when you get what is commonly called the action groupoid. In this case the morphism $p: \Gamma \rtimes G \to G$ is known as a covering morphism of groupoids, and all covering morphisms of $G$ arise in this way.

I feel the use of covering morphisms of groupoids makes for a nice exposition, base point free, of the theory of covering spaces. Such an idea was pointed out for the simplicial case in the 1967 book on simplicial theory by Gabriel and Zisman, was used in the first 1968 edition of my book, and is used in Peter May's 1999 book "A concise course in algebrac topology".

Update: I should also add that the notion of action of a groupoid on a groupoid is given, following Ehresmann, in my paper

[11] Groupoids as coefficients'', Proc. London Math Soc. (3) 25 (1972) 413-426.

available here. The aim was cohomology with coefficients in a groupoid. One of the methods exploited is fibrations of grouopids.

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