# 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.

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June 9, 2007 -

## 15 Comments »

1. Is that best example possibly the subject of my highschooler project? :)

Comment by michiexile | June 9, 2007 | Reply

2. If you’re referring to the fundamental groupoid of a space, yes. Of course, I haven’t even defined a topology yet, so that’s a ways off.

Comment by John Armstrong | June 11, 2007 | Reply

3. You know, if the places where I first read about category theory explained it this clearly, I wouldn’t hate it nearly so much as I do now.

Comment by Alon Levy | June 11, 2007 | Reply

4. Alon, I’m glad I could make it a bit clearer, then.

I think the big failing, even in books like MacLane’s (/me bows reverently), is not taking specific examples the audience already knows in another language and working them through in full gory detail like this. Sure, some who are inclined to categories can figure out how to fill in the details, but others who aren’t given to the subject will find it dense and impenetrable.

Incidentally, I’d say the same goes for algebraic geometry, where I find myself on the other side of the book.

Comment by John Armstrong | June 11, 2007 | Reply

5. I haven’t even looked at that book… the only category theory book I’ve picked up is Categories for the Working Mathematician. I’m talking mostly about verbal explanations, things I’ve read online (e.g. on Wikipedia and on GM/BM), and even the functorial constructions introduced in Hartshorne.

Comment by Alon Levy | June 12, 2007 | Reply

6. Sorry.. CWM is MacLane’s book. The reverence is.. partly an inheritance from my advisor. MacLane preceded him as a Yale undergraduate by a few years and held (may still hold) the record for undergraduate GPA. His book is near-legendary, and he was an excellent mathematician besides, so the joking reverence is not exactly unearned.

I get the category theory in Hartshorne. What I don’t ever quite get (and I’m reading through it again right now) is what any of it is supposed to be good for. I have yet to see a treatment of the jump Hartshorne makes between chapters 1 and 2 — and which all other authors are obliged to make, to be sure — which really motivates it to my satisfaction. Maybe some day I’ll run across an expository mathematical blogger with a liking of algebraic geometry and I will be enlightened, but for now there’s just something that eludes me about the subject.

Comment by John Armstrong | June 12, 2007 | Reply

7. With my background, it feels continually weird to see CWM refered to as “MacLane’s book” without qualification.

I’m used to having MacLane and Hilton&Stammbach being the house bibles, with Weibel as the catechism.

But then again, I’m one of those really weird people who learned category theory to grasp homological algebra, and who learnt algebraic topology -after- getting to grips with homological algebra…

Topology just kept getting easier once it got algebraic though.

Comment by michiexile | June 12, 2007 | Reply

8. My understanding is that you need to rigorously define the pushforward and pullback for certain things about schemes to work properly. I don’t exactly remember which because I made a conscious effort to forget anything that had to do with categories, but I distinctly remember how sheaves of modules require a fairly evolved understanding of one of the two constructions.

Comment by Alon Levy | June 13, 2007 | Reply

9. Oh that much is easy. I get sheaves, because sheaves are functors :D. Pullback should be easy, and pushout should be hard. Limits of sheaves are automatically sheaves, but colimits of sheaves may only be presheaves, so you need to sheafify at the end.

Actually, I’ll be doing all the category theory you need for this soon enough, and the application sheaves once I have a bit of topology under my belt.

Comment by John Armstrong | June 13, 2007 | Reply

10. In the 9th paragraph at the end, shouldn’t it be “with no morphisms between ‘any’ distinct objects.” not “with no morphisms between ‘and’ distinct objects.” Very clear exposition, I had just picked up a book on Lie groupoids and was just dreading having to diagram what they were talking about in their description of groupoids. Thanks very much.

Comment by Ralph Gardner | November 19, 2007 | Reply

11. Glad to help. Thanks for catching the typo.

Comment by John Armstrong | November 19, 2007 | Reply

12. […] any category and throw away all the morphisms that aren’t invertible. We’re left with a groupoid, and like any groupoid it falls apart into a bunch of “connected” pieces: the […]

Pingback by Isomorphisms of Vector Spaces « The Unapologetic Mathematician | October 17, 2008 | Reply

13. You might like to look at my groupoid page at

http://www.bangor.ac.uk/r.brown/gpdsweb.html

which gives links to the history of groupoids, in particular the important geometric work of Charles Ehresmann.

I came into groupoids in 1965 via the fundamental groupoid, and Higgins’ work applying groupoids to group theory. He used the action groupoid of a group G on a set of cosets to give a “covering morphism” p: H \to G of groupoids, and then lifted presentations of G to presentations of H, which could be analysed geometrically.

I rather like the notion of the action groupoid for a group G operating on a set S to be seen as a semidirect product, since one then uses the same notation for G operating on a groupoid, which is relevant to orbit space considerations.

Higgins and I found in 1974 that groupoids, unlike groups, had good higher dimensional applications in algebraic topology, using higher homotopy groupoids. It is a long story though.

Ronnie Brown

Comment by Ronnie Brown | March 6, 2009 | Reply

14. It looks interesting, thanks. Something to check out while I’m on break next week.

Comment by John Armstrong | March 6, 2009 | Reply

15. […] map that, given , returns every such that ; we’ll call this as a nod to the abstract perspective, so that . In the dihedral group, there are always two ways to get from to : one can either rotate […]

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