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A Bicategory of Groupoids July 23, 2007

Posted by Chris Schommer-Pries in Category Theory, groupoids, topology.
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I want to talk about an interesting 2-category of topological groupoids that I’ve been thinking about recently. Let’s start with the basics; what is a groupoid? Well a group can be thought of as a category with one object and with every morphism invertible. A groupoid is the same thing, except that there can be multiple objects. Why are these interesting? well groupoids generalize three different notions at once: sets, equivalence relations, and groups. We’ve seen how groups enter. Equivalence relations on sets can also be viewed as categories where there are exactly one or zero morphisms between two objects: one if the objects are equivalent, zero otherwise.

Obvious Fact: Since groupoids are categories we can talk about functors between groupoids and natural transformations between these. Hence groupoids form a 2-category.

What happens if we add topology? What if we want a space of objects and a space of morphisms? (source, target and identiy maps are countinuous, of course). Well things start to break down.

The main problem is this: suppose that we have a space that is the union of two open subspaces,

X = U_1 \cup U_2

For notation, let U_{ij} be the intersection of U_i and U_j.

Then we can form two topological groupoids out of this:

  • X viewed as a space of objects with only the identity morphisms.
  • U, which has objects the disjoint union of U_1 and U_2, and where the morphisms are the disjoint union of U_{11}, U_{12}, U_{21} and U_{22}. The source and target maps are the obvious inclusions.

U is a special case of an equivalence relation: two points in U_1 and U_2 are equivalent if they correspond to the same point in X.

Moreover there is a natural map/functor from U to X which sends a point of U to the corresponding point in X.

If we forget the topology, then this is an equivalence of categories/groupoids, and morally we want this to hold if we have topology around too. These two topological categories should be equivalent.

Therein lies the problem. If we keep the topology around (and require our functors to be continuous) then the obvious map is not in general an equivalence. For example suppose that X is connected. Then to have the “inverse” of the natural functor from U to X, we would need a functor from X to U which is surjective on equivalence classes of objects. As long as neither U_i is all of X, then this is impossible since we would need a continuous map from a connected space (X) whose image lies in distict components of a disconnected space (the objects of U). No such map can exist.

What are we to do?

Well fortunately there is a pretty nice solution. The idea is to replace functors with “bibundles”. First, let’s back up a little and talk about ordinary groupoids again.

When I said that groupoids generalize spaces and groups, I did what every category theorist would consider a big no-no; I talked about the objects and said nothing about the morphisms. Since it’s Monday and I’m lazy I’ll leave it as a exercise to check that if you consider two sets as groupoids then the functors between them are the same as just maps of sets and the natural transformations are identities. Similarly if you take two groups and view them as groupoids, then functors between them correspond to homomorphisms between the groups (natural trans. are more interesting here).

Now I’m going to start tweaking our 2-category of groupoids until we have something that will work for topological groupoids.

Let’s set some notation. Groupoids will be denoted as pairs (G_0, G_1), where G_0 is the set of objects and G_1 is the set of morphisms. All of the structure maps are going to be implicit. Thus if S is a set then (S,S) is S viewed as a groupoid with only identity morphisms and if G is a group, (pt, G) is G viewed as a groupoid.

There is an important definition which we’ll need: a right action of a groupoid (G_0, G_1) os a set S consists of a map S \rightarrow G_0, together with an action map S \times_{G_0} G_1 \rightarrow S, compatible with the composition of (G_0, G_1).

This is sort of difficult to visualize at first, so let’s do some examples.

If our groupoid is just an ordinary group, then G_0 is just a point. Thus the first map is no data. The second map just becomes an ordinary action map. So we see in this case it is just the usual notion of an action of G on S.

If our groupoid is a set (Y,Y), then the first map is a map of sets from S to Y, but the second map becomes trivial. It reduces to a map from S to S, but since it’s “compatible with composition” it must be the identity map.

Of course we define left actions in a totally analogous way. If we have two groupoids then we can define biactions on a set to be a left action of one groupoid and a right action of the other which commute. There are also obvious notions of “equivariant maps” of sets with groupoid actions.

Now we can use this to “change” our 2-category of groupoids in a sneaky way. We will replace functors with certain sets with biactions.

Let’s consider the case where we have two groups, G and H, which we are considering as groupoids. Note that the set H has a natural right action by the groupoid H (the map to the point is for free and the action map is also for free). Notice that the group H acts on this set H from the left as well, and that this commutes with the right action. Thus we can turn the set H into a set with a G-H biaction. The map to the objects of G is again free and G acts on the left of H via the map of groups from G to H.

Summarizing: Out of a functor between groups we can construct a set with a G-H biaction, and such that H acts freely and transitively on the fiber over G_0 = pt. We’ll call this set a bibundle for reason that will be obvious later.

Exercises: Show that there is a natural notion of composition of bibundles (between groups) and that it matches up with the composition of functors. Also show that natural transformations induce equivariant maps of bibundles.

Similarly suppose that we have two sets, X and Y, viewed as groupoids and a functor f between them. We have a similar story where we can build a space P which has a left X action and a right Y action. Recall that this is just a map to the set X and a map to the set Y. Let’s take P=X, with the identity map to X and the map f from P to Y. Note that the fibers of the map to the objects of the goupoid X are just single elements of P. But since Y is just a space, it acts “free and transitively” on these fibers. The key new ingrediant that we have from this example is that the map from P to (the objects of) X is surjective.

Definition: A right principle bibundle between two groupoids G and H is a set P with a G-H biaction such that the map from P to the objects of G is surjective and such that the fibers are acted on by H freely and transitively.

This extends the above construction from groups to spaces and in fact we can extend it to all groupoids by taking
P = G_0 \times_{H_0} H_1,
where the map from G_0 to H_0 is the functor on objects. The value of the functor on morphisms yields the action on P.

Exercise: Show that this construction yields a (weak) 2-functor from the 2-category of (groupoids, functors, nat. trans) to the bicategory of (groupoids, bibndles, equivariant maps). Show that this is an equivalence of bicategories.

Whew! what a detour, eh? Well now we are all set to fix the category of topological groupoids. If we return to the bad example above where X is the union of two open sets, we see that the problem that we ran into was a local one. If we could shrink X enough (say to one of the open sets U_i) the we can get an equivalence of topological categories (the shrunken X is equivalent to its preimage category in U).

Based on our bibundle ideas we define a new bicategory where the 1-morphisms are topological bibundles (all maps are continuous). It turns out that this includes too many morphisms so we also require that our 1-morphisms are locally trivial (in the source). This means that if we shrinch the source groupoid enough, then the bibundle becomes isomorphic to one of the form
P = G_0 \times_{H_0} H_1.

I think this is a really neat bicategory. First of all, it fixes the problem we had before, U and X become equivalent objects in this bicategory. It is a place where things which are “locally a functor” make sense. It also includes lots of interesting geometric stuff. For example, suppose that G is a topological group and X is a topological space. What is the hom category hom(X, G)?

It’s the category of right G-principle bundles over X! Wow!

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Comments

1. John Armstrong - July 23, 2007

At a first glance, this “bibundle” seems to be an anafunctor from one groupoid to another, right?

2. Robin - July 23, 2007

Bibundles?

Pity the poor category theorists. We already have to contend with the fact that these gadgets are variously known as profunctors, distributors, modules, and bimodules. The last thing we need is yet another name for them! :-)

3. Chris Schommer-Pries - July 24, 2007

I’m not an expert on anafunctors, profunctors, distributors, etc. But from what I’ve been able to gleam from the net, these bibundles are not (obviously) any of these things.

Remeber that this bibundle bicategory depends on the topology of the groupoids. If you forget the topology you get something equivalent to (groupoids, functors, nat. trans.). Anafunctors and profunctors=distributors=etc are meant to generalize functors and natural transformations.

You might be able to discuss this bicategory using these concepts if you replace your topological groupoid with something like its “category of open sets” (= site). I say “something like” because really this would have to be 2-categorical in nature, but even then I’m not sure you actually need the technical machinery of anafuctors or profunctors.

4. Robin - July 24, 2007

Perhaps I have misunderstood your construction. I was referring to the bicategory that you get without restricting to locally trivial 1-morphisms. Isn’t that the bicategory of internal groupoids and internal profunctors (internal to topological spaces)?

(Internal profunctors don’t seem to be terribly well known. I only learned about them recently, from Todd Trimble. Apparently they are discussed in Johnstone’s Topos Theory, and they’re also mentioned in volume 1 of Borceux’s handbook (exercise 8.4.3).)

I’m sorry if this is wide of the mark.

5. Chris Schommer-Pries - July 24, 2007

Ok let’s forget the “locally trivial” part. It’s not so important anyway.

I totally agree that the objects are just internal groupoids.

I don’t have either of your references handy, but wikipedia says a profunctor from categories C to D is a functor

F: C \times D^{op} \rightarrow Set

If this is correct, I’m guessing that an internal profunctor maps to Top? That’s just a guess. Please correct me if I’m wrong, I don’t know very much about profunctors.

If there is a way to turn one of these into a bibundle, I’d really like to know. I’m not seeing it right away. Somehow we need to get a space P with maps to the objects of C and D… hmm…

6. Robin - July 24, 2007

Okay, so let \mathbb{C} and \mathbb{D} be topological categories, i.e. internal categories in \mathrm{Top}. Morally, a profunctor ought to be an internal functor \mathbb{C}\times\mathbb{D}\to\mathrm{Top}. But you can’t say that directly, since \mathrm{Top} itself is not a topological category (for size reasons, if nothing else). Therefore a bit of subterfuge is required.

There is a general notion of “base-valued functor” on an internal category \mathbb{C}, which captures the informal idea of an internal functor \mathbb{C}\to\mathrm{Top}. It consists of an object P\in\mathrm{Top} together with a morphism p_0: P\to \mathbb{C}_0 in \mathrm{Top}; and an arrow p_1: \mathbb{C}_1\times_{\mathbb{C}_0}P\to P in \mathrm{Top}, satisfying some axioms.

In the case of an internal profunctor, you get a space P together with an arrow p_0: P\to\mathbb{C}_0 \times \mathbb{D}_0, which does indeed give you a space equipped with maps to the objects of \mathbb{C} and \mathbb{D}.

So it’s looking plausible, at least?

7. Robin - July 24, 2007

Correction: I accidentally omitted an {}^\mathrm{op} in the second sentence.

8. Robin - July 24, 2007

Perhaps to clarify a bit, I am claiming that a left action (of a topological groupoid, or more generally a topological category, on a space) is a special case of the general notion of internal base-valued functor, and that a topological bibundle is a special case of the general notion of internal profunctor.

Furthermore, almost all of the theory of profunctors can be reproduced in the internal setting. In particular, every (internal) functor gives rise to an adjoint pair of profunctors (one of which Chris discussed above).

9. Joel Kamnitzer - July 24, 2007

Is it true that this bibundle 2-category is equivalent to the 2-category of topological stacks?

I think that this is explained in this paper

http://arxiv.org/PS_cache/math/pdf/0306/0306176v1.pdf

by David Metzler. I believe that what you call bibundle, he calls Hilsum-Skandalis morphism.

10. Urs Schreiber - July 25, 2007

what you call bibundle, he calls Hilsum-Skandalis morphism.

Which, in turn, is apparently indeed the same as a (saturated) anafunctor, or at least that’s what Toby Bartels told me.

11. John Armstrong - July 25, 2007

I knew it! Thanks, Urs.

12. Chris Schommer-Pries - July 26, 2007

Hold on Urs,

So yes these bibundles are also called Hilsum-Skandalis morphism (at least I know they are also called this when you are working with smooth manifolds, not just top. spaces).

However, latter on (in your same link) Toby says (regarding whether Hilsum-Skandalis morphism are the same thing as saturated anafunctors)

I’m not sure that this is really true. But something like it is.

So it still doesn’t seem settled. Perhaps you could explain the relationship in more detail?

Also thanks, Robin, for your comments. It seems plausible that special knids of profunctors are equivalent to Hilsum-Skandalis morphisms, but that profunctors are actually more general. This would be consistent with Hilsum-Skandalis morphisms being equivalent to anafunctors, since anafunctors (acording to Urs’ link) are a specail case of profunctors (they have to be “representable”).

13. Chris Schommer-Pries - July 26, 2007

Joel,

That’s an interesting question that I’m not completely sure of the answer. I’ll tell you what I know and maybe we can piece it together. (btw, this is all contained in that paper you referenced by David Metzler).

First of all, there is a 2-functor from this bibundle 2-category to the 2-category of topological stacks. This is totally analogous to the Yonneda functor which embeds topological spaces into the category of sheaves on the big topological site. A groupoid is sent to the (lax} functor from spaces to categories, which sends a space X to the category hom(X,G). This (lax) functor is the stack associated to G.

Furthermore, two groupoids get sent to equivalent stacks if and only if they are equivalent in this bibundle 2-category. I believe that any morphism between the image stacks can be realized as a morphism between the orginal groupoids, or more precisely that this Yonneda-type 2-functor is an equivalence on hom categories. (I think this becomes clear if you work it out by hand).

If true, then this 2-functor should be thought of as fully faithful.

No a sticky issue… What do we mean by stack? In the world of algebraic geometry we almost never mean any stack. We usually mean something like “algebraic” stack. This condition implies that the stack is “locally representable” in a certain technical sense. If we insist on the analogous condition for topological or smooth stacks then it is true that they are equivalent to ones coming from a groupoid.

So our functor is also essentially surjective. Now I don’t know what to do. If we just had ordinary 1-categories, we’d be done since a fully faithfull essentially surjective functor is an equivalence by the axiom of choice. In 2-category land there are all sorts of pitfalls with the axiom of choice, so it is not clear if they are actually equivalent.

If anyone out there knows how to clear this up, I’d much appriciate it. Incidentally, anafunctors were invented for exactly this reason. They generalize functors in just the right way so you can avoid problems like the above. You don’t have to rely on the axiom of choice. I’m almost 100% sure that these two 2-categories are equivalent via “2-anafunctors”.

What’s the right terminology? 2-anaequivalent?

Finally, you can probably see that I actually prefer this 2-category to the 2-category of stacks. I feel that these bibundles are more geometric and that the stack language is more like viewing a space as a sheaf on a certain site; it’s more removed from the geometry.

Hope this helps.

14. Urs Schreiber - July 27, 2007

I’m not sure that this is really true. But something like it is.

Right. So let’s be careful. I haven’t sat down and thought this through in detail. Somebody should.

15. David Roberts - August 9, 2007

Dorette Pronk showed in 1996 that topological groupoids which are tame (so the topological spaces involved are) form a bicategory which admits a calculus of fractions where we invert the internal functors which are fully faithful and essentially surjective (these conditions can be written internally – it’s a cute exercise). Bibundles are the same as a span of groupoids such that the right `foot’ is the quotient of the middle groupoid by the action arising using the left leg as moment map (and vice versa). This condition is the same as both legs being fully faithful and essentially surjective. These spans are then the equivalences. In general morphisms are described not by bibundles, but by bundles, which correspond to spans where the first leg is ff and ess surj.

The bicategory of fractions in this case is equivalent to topological stacks, although Pronk uses the word etendu, I think to avoid confusion with the usual notions of stack (Deligne-Mumford etc).

16. Urs Schreiber - August 10, 2007

In general morphisms are described not by bibundles, but by bundles, which correspond to spans where the first leg is ff and ess surj.

And such a span is an anafunctor! :-)


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