The Unapologetic Mathematician

Groupoids (and more group actions)

A groupoid is the natural extension of a group, considered as a category. A groupoid is a category where each morphism is invertible. If a groupoid has only one object it’s a group.

These structures are the central players in John Baez’ “Tale of Groupoidification”, which started in this issue of This Week’s Finds. They’re also a rather popular subject over at The n-Category Café.

Unfortunately, one of the best examples of a groupoid is still beyond us at this point, but there are others. One nice place groupoids come up is from group actions. In fact, Baez is of the opinion that the groupoid viewpoint is actually more natural than the group action viewpoint, but I think it would be difficult to start teaching from that end.

So how does this work? Let’s say we have a group $G$ and a left action of $G$ on a set $S$. Now let’s build a groupoid from this information. The high-level view is that every single move from one state (element of $S$) to another will be a morphism, and we can compose morphisms if the outgoing state of the first is the incoming state of the second. Now for the details.

We take the objects of our groupoid to be the set $S$, and the morphisms to be the set $G\times S$. The source of a morphism $(g,s)$ will be $s$, and the target will be $g\cdot s$. For each object $s$ we have the identity morphism $(e,s)$, where $e$ is the identity of $G$. If we have two morphisms $(g,s)$ and $(g',s')$, and $g\cdot s=s'$ (so the pair are composable), then we have the composite $(g',s')\circ(g,s)=(g'g,s)$. Note that this still has source $s$, and its target is $(g'g)\cdot s=g'\cdot(g\cdot s)=g'\cdot s'$, as it should be.

Well this is enough to show we have a category, but now we need to show that each morphism has an inverse. If we have a morphism $(g,s)$ going from $s$ to $g\cdot s$, we’ll need the inverse to go from $g\cdot s$ to $s=(g^{-1}g)\cdot s=g^{-1}\cdot(g\cdot s)$. The clear choice is $(g^{-1},g\cdot s)$, and indeed we find that $(g^{-1},g\cdot s)\circ(g,s)=(e,s)$, and $(g,s)\circ(g^{-1},g\cdot s)=(e,g\cdot s)$. Thus we have a groupoid.

This result we call the “action groupoid” or the “weak quotient”, and write $S//G$. Remember that the “real” quotient is the set of orbits of $G$ on $S$ — we consider two elements of $S$ to be “the same” if they are related by an element of $G$. Here we don’t consider them to be the same; we just make them isomorphic. Since we’ve replaced “the same” by “isomorphic”, we call this a “weak” quotient.

Now, an action groupoid is not just any groupoid. For one thing, we’ve got a functor $S//G\rightarrow G$. Just send every single object of $S//G$ to the unique object of $G$, considered as a category. Then send the morphism $(g,s)$ to $g$. We’re sort of folding up the groupoid $S//G$ into the groupoid (with one object) $G$. But here’s the thing: this functor is faithful! Indeed, we see that $\hom_{S//G}(s,s')$ is the set of $g$ so that $s'=g\cdot s$. In particular, it’s a subset of the elements of $G$, and the functor acts on this hom set by the inclusion function of this subset into $G$, which is injective. Even though the functor loses a lot of information (like, which state a given morphism started at) it’s still faithful because “faithful” just pays attention to what happens in each hom set, not what happens to the objects.

It turns out we can find a (somewhat) simpler groupoid $\mathcal{H}$ equivalent to $S//G$ and compatible with this faithful functor. That is, if the functor above is $F:S//G\rightarrow G$, we can find a groupoid $\mathcal{H}$ with functors $E:\mathcal{H}\rightarrow S//G$ and $I:\mathcal{H}\rightarrow G$. The functor $E$ will be an equivalence, $I$ will be faithful, and we’ll have $I=F\circ E$. The really special thing about $H$ is that it will be a collection of groups with no morphisms between any distinct objects.

For the objects of $\mathcal{H}$, take the set of $G$-orbits of $S$. There will be no morphisms between distinct objects, so we just need to specify a group of morphisms on each object. So, given an orbit $O\subseteq S$ pick a point $x\in O$ and let $\hom_\mathcal{H}(O,O)=G_x$ — the stabilizer of $x$. Of course, different representative points may give different stabilizers, but the stabilizers of any two points in the same orbit are isomorphic. For the functor $I$, just send each object of $\mathcal{H}$ to the single object of $G$ and include each hom set into $G$ as a subgroup, just like we did for $F$.

Now we define our equivalence. Since we’ve already picked a representative point in each orbit, let’s just send the object of $\mathcal{H}$ corresponding to that orbit to the object of $S//G$ corresponding to that point. Then we can just send $\hom_\mathcal{H}(O,O)=G_x=\hom_{S//G}(x,x)$ to itself. Clearly this is fully faithful, and it’s also essentially surjective because every point of $S$ is in some $G$-orbit. So we have an equivalence. Also it should be clear that $I=F\circ E$.

In fact, we can adapt this to any groupoid $\mathcal{G}$ with a faithful functor to a group $G$. Just replace “in the same orbit” by “have an arrow between them”. Then chop up the objects of $\mathcal{G}$ into equivalence classes, make each one an object of our new groupoid $\mathcal{H}$, and make the morphisms on an object in $\mathcal{H}$ the “stabilizer” of a representative object from $\mathcal{G}$. It all goes through to give the same sort of factorization.

And then if you push just a little bit harder you can take any discrete groupoid $\mathcal{H}$ (“discrete” = no morphisms between distinct objects) with a faithful functor $\mathcal{H}\rightarrow G$ and puff it up into an action groupoid. In fact, it suffices to show how to do this for a single object in $\mathcal{H}$, since all the others go the same way.

In this case we have a subgroup $H\subseteq G$ and we need to see it as the stabilizer of some $G$ action. We use the set $G/H$ of left cosets of $H$ in $G$. This may not be a group (since $H$ may not be normal), but it’s at least a set, and $G$ certainly acts on it by left-multiplication. What subgroup of $G$ fixes the coset $eH$? Exactly $H$. So, we puff $H$ out into $G/H$ and take its weak quotient by $G$ to get an action groupoid equivalent to $H$.

Putting this all together we see that any groupoid $\mathcal{G}$ with a faithful functor $\mathcal{G}\rightarrow G$ is equivalent to the action groupoid $S//G$ for some $G$-set $S$. And any $G$ action gives an action groupoid with a faithful functor to $G$. The two concepts are essentially the same.

June 9, 2007