Category Theory
Zulip Server
Archive

You're reading the public-facing archive of the Category Theory Zulip server.
To join the server you need an invite. Anybody can get an invite by contacting Matteo Capucci at name dot surname at gmail dot com.
For all things related to this archive refer to the same person.

Stream: learning: questions

Topic: groupoids with two objects

David Egolf (Feb 15 2022 at 18:52):

A group is a groupoid with a single object. People have put a lot of work in to understanding groups! What about groupoids with two objects? Have people studied those quite a bit as well? (Or does somehow their study reduce to the study of groups?) I know groups are important in physics. Are groupoids with two objects an important topic in physics as well?

Joe Moeller (Feb 15 2022 at 18:54):

There are two possibilities. Either there exists a map between the two objects, they're isomorphic, and their automorphism groups are the same, or there isn't a map, in which case it's nothing more than a pair of groups.

Mike Shulman (Feb 15 2022 at 18:59):

However, if there exists a map but no distinguished map, then there is no specified isomorphism between the two groups. Although there is a specified isomorphism-modulo-conjugation.

Mike Shulman (Feb 15 2022 at 18:59):

I do think this is a case where there's not much to say about the $n=2$ case that doesn't apply equally well to all $n>1$.

David Egolf (Feb 15 2022 at 19:03):

Mike Shulman said:

However, if there exists a map but no distinguished map, then there is no specified isomorphism between the two groups. Although there is a specified isomorphism-modulo-conjugation.

What does it mean for there to be a "specified" isomorphism? I guess you mean that if there are multiple isomorphisms between the two groups, then the two groups are isomorphic to eachother in a number of different ways, with no one way being distinguished.

I've heard about conjugation, but I don't know what it means for an isomorphism to be specified "modulo conjugation".

Mike Shulman (Feb 15 2022 at 19:04):

One way to say what I mean is that there is no function from "two-object groupoids" to "pairs of two groups with an isomorphism between them" that you can define without using the axiom of choice.

Mike Shulman (Feb 15 2022 at 19:05):

But there is a function to "pairs of two groups $G,H$ with an element of $Iso(G,H)/H$", where $H$ acts on $Iso(G,H)$ by conjugation, $(h,\phi) \mapsto (g \mapsto h^{-1}\cdot \phi(g) \cdot h)$.

David Egolf (Feb 15 2022 at 19:14):

I think I sort of follow (?). You start with a groupoid having two objects, and you want to associate it with a pair of groups with an isomorphism between them. However, given such a groupoid, there could be multiple ways to create a pair of groups with an isomorphism between them that could be reasonably associated with that groupoid.
However, if we consider two isomorphisms between the groups to be basically the same if one can be obtained from the other when $H$ acts on one by conjugation, then we can send a groupoid to a pair of groups together with an equivalence class of isomorphisms between them (with the equivalence relationship induced by the conjugating action of $H$ on $Iso(G,H)$).

David Egolf (Feb 15 2022 at 19:16):

If that's right, then a groupoid with two objects is "just" a pair of isomorphic groups together with a collection of isomorphisms between them related by conjugation (?).

Mike Shulman (Feb 15 2022 at 19:16):

Yes, that's right. If $G = Aut(x)$ and $H=Aut(y)$ in your groupoid, then any morphism $f:x\to y$ yields an isomorphism $\phi_f : G\cong H$ by $\phi_f(g) = f\circ g \circ f^{-1}$. A different $f':x\to y$ yields a different isomorphism $\phi_{f'}$, but they are related by conjugation by $f\circ (f')^{-1}\in H$.

Mike Shulman (Feb 15 2022 at 19:17):

I stopped short of saying "just" because I couldn't think of how to construct a 2-object groupoid from a pair of groups and a conjugation-equvialence-class of isomorphisms between them. I don't remember whether I should expect this to be possible or whether there's more data there too.

Mike Shulman (Feb 15 2022 at 19:17):

(Also, of course, this is all about connected 2-object groupoids. As Joe said, a disconnected 2-object groupoid is just a pair of unrelated groups.)

John Baez (Feb 15 2022 at 19:52):

David Egolf said:

Mike Shulman said:

However, if there exists a map but no distinguished map, then there is no specified isomorphism between the two groups. Although there is a specified isomorphism-modulo-conjugation.

What does it mean for there to be a "specified" isomorphism? I guess you mean that if there are multiple isomorphisms between the two groups, then the two groups are isomorphic to each other in a number of different ways, with no one way being distinguished.

Exactly - in this context "distinguished" implies "specified". I could define exactly what "distinguished" means, but for now I'll just say it does not mean "having a beard and monocle, and wearing a tuxedo".

John Baez (Feb 15 2022 at 19:53):

In general sometimes you can "specify" something in the process of a mathematical construction, and from then on it's "specified".

John Baez (Feb 15 2022 at 19:54):

Sometimes people use "distinguished" to mean something that's inherently specified, not by an arbitrary act of the mathematician, but in its very nature.

John Baez (Feb 15 2022 at 19:56):

For example if we just say $X$ is a set with 7 elements, it has no distinguished element. We can specify an element and call it $x$. Then we have a set with a specified element.

But if we say $X$ is a group, then it has a distinguished element, $1$. And if all we know is that $X$ is a group, this is the only distinguished element.

John Baez (Feb 15 2022 at 19:57):

All this can be made precise using category theory or type theory - I leave that as an exercise to the readers of this Zulip. It takes precise definitions to prove actual theorems about these concepts.

Mike Shulman (Feb 15 2022 at 19:58):

John Baez said:

We can specify an element and call it $x$. Then we have a group with a specified element.

I think you mean a set with a specified element. (A.k.a. a "pointed set".)

John Baez (Feb 15 2022 at 20:08):

Yes. I went back and fixed that.

Oscar Cunningham (Feb 16 2022 at 10:22):

There is more than one possibility for what the morphisms in the category of groupoids with two objects would be. Are we allowed to send both objects to the same object in the destination?

Zhen Lin Low (Feb 16 2022 at 11:15):

I think it is helpful to distinguish between "groupoids with two specified objects" and "groupoids that merely have two (distinct) objects". The natural choice of morphism for the former would have to be bijective on objects (and not allowed to swap the two objects either).

John Baez (Feb 16 2022 at 15:39):

I think we have been talking - implicitly - about the (2-)category of groupoids $X$ with two specified objects $x$ and $y$, say ordered triples $(X,x,y)$ where $x$ is the first object and $y$ is the second object. A morphism in this category is a functor sending the first object to the first object and the second to the second. (A 2-morphism would be a natural transformation, necessarily a natural isomorphism.)

Oscar Cunningham (Feb 16 2022 at 16:16):

To make one-object-groupoids into groups we have to add the extra condition that the natural transformations are the identity at the specified object. Otherwise we get that two conjugate group homomorphisms are isomorphic. So likewise I think we should demand that the natural transformations in the two object case are the identity on both of the specified objects.

Martti Karvonen (Feb 16 2022 at 16:32):

That's probably right, but sometimes it does make sense to think of groups as a 2-category with natural transformations/conjugations as the 2-cells. For instance, HNN-extensions are coinserters in this 2-category.

David Egolf (Feb 16 2022 at 16:34):

John Baez said:

I think we have been talking - implicitly - about the (2-)category of groupoids $X$ with two specified objects $x$ and $y$, say ordered triples $(X,x,y)$ where $x$ is the first object and $y$ is the second object. A morphism in this category is a functor sending the first object to the first object and the second to the second. (A 2-morphism would be a natural transformation, necessarily a natural isomorphism.)

Hmmm. I was intending to ask about categories that have exactly two objects, and where every morphism is an isomorphism. I don't know enough about 2-categories to say if what you're describing here is equivalent to that in some sense.
I guess @John Baez is here talking about morphisms between groupoids having two distinguished objects, now that I read what he said more carefully. Ohhh... and now I see he was responding to Oscar Cunningham. Sometimes it just takes me a while to figure out what you all are saying!
It makes sense that to better understand groupoids with two objects, we'd want to consider morphisms between them!

David Egolf (Feb 16 2022 at 16:40):

Oscar Cunningham said:

To make one-object-groupoids into groups we have to add the extra condition that the natural transformations are the identity at the specified object. Otherwise we get that two conjugate group homomorphisms are isomorphic. So likewise I think we should demand that the natural transformations in the two object case are the identity on both of the specified objects.

I thought a one-object-groupoid was a group? I think this comment is maybe talking about a 2-category of one-object-groupoids, somehow(?). If someone would be willing to explain this a little, I'd appreciate it.

Oscar Cunningham (Feb 16 2022 at 16:46):

The problem is that we usually consider groupoids to form a 2-category, with groupoids as objects, functors as morphisms, and natural transformations as 2-morphisms. So if we naively restrict our attention to groupoids with one object, then we get a 2-category with groups as objects, homomorphisms as morphisms, and conjugations as 2-morphisms. But what we usually want is to consider groups as forming an ordinary (1-)category. So we need to impose some extra condition to kill the unwanted 2-morphisms. For example demanding that they be the identity at the single object.

David Egolf (Feb 16 2022 at 16:51):

Oscar Cunningham said:

The problem is that we usually consider groupoids to form a 2-category, with groupoids as objects, functors as morphisms, and natural transformations as 2-morphisms. So if we naively restrict our attention to groupoids with one object, then we get a 2-category with groups as objects, homomorphisms as morphisms, and conjugations as 2-morphisms. But what we usually want is to consider groups as forming an ordinary (1-)category. So we need to impose some extra condition to kill the unwanted 2-morphisms. For example demanding that they be the identity at the single object.

Thanks for explaining! I had learned about conjugation in an introductory abstract algebra book, but learning about its connection to natural transformations is super cool!
Why is it necessary to "kill" the 2-morphisms? Why can't one just truncate their attention and say, "Yes, we can form a 2-category, but we just want a 1-category at the moment so ignore these things that could be 2-morphisms."
I suppose what's appropriate just depends on the context.

Oscar Cunningham (Feb 16 2022 at 17:04):

You can, but it's 'evil' to do so because it takes two isomorphic things (conjugate homomorphisms) and sends them to things which aren't isomorphic (different homomorphisms). Good categorical constructions should respect isomorphisms.

Mike Shulman (Feb 16 2022 at 17:13):

This actually gets back to what John said a while ago:
John Baez said:

For example if we just say $X$ is a set with 7 elements, it has no distinguished element. We can specify an element and call it $x$. Then we have a set with a specified element.

It's not correct to say that "a group is a groupoid with one object" if we mean by that that its set of objects has the property of "having one element". This is because the collection of such groupoids naturally forms a 2-category, which is equivalent to the 2-category whose objects are groups (in the ordinary sense), whose morphisms are homomorphisms, and whose 2-cells are conjugations. In particular, two groups can be equivalent in this 2-category if they are not isomorphic as groups, so "a group" (which, in practical mathematics, always implies "considered up to isomorphism") is not the same as an object of this 2-category.

What is correct is to say that "a group is a groupoid with one specified object". In general, saying that something is specified implies that it has to be preserved by the morphisms. So for instance, the category of pointed sets (sets with a specified element) has morphisms being functions that preserve the specified elements: $f:(X,x) \to (Y,y)$ is a function $f:X\to Y$ such that $f(x)=y$. Of course, when both domain and codomain have exactly one element, all functions must preserve those elements, so "the category of one-element sets" is the same no matter whether or not the elements are specified.

But things are different for one-object groupoids. A functor between one-object groupoids is the same regardless of whether the one objects are specified or not: it's just a group homomorphism between the automorphism groups. But a transformation between functors $f,g:G\to H$ involves a function ${\rm ob}(G) \to {\rm mor}(H)$ giving the components of the transformation, and even if ${\rm ob}(G)$ is a one-element set, the same is not true of ${\rm mor}(H)$. However, if $\star$ is the one object of $H$, then $H(\star)$ does have a distinguished element, namely the identity ${\rm id}_\star$. Thus, a transformation between pointed one-object groupoids must be the identity, and so the 2-category of pointed one-object groupoids is equivalent to the 1-category of groups.

David Egolf (Feb 16 2022 at 17:18):

Thanks so much for the detailed explanation! It's going to take me a bit of time and effort to absorb it.

Oscar Cunningham (Feb 16 2022 at 17:33):

In particular, two groups can be equivalent in this 2-category if they are not isomorphic as groups

What's an example of this?

Mike Shulman (Feb 16 2022 at 17:34):

Oh wait, sorry, I think I messed that up.

Mike Shulman (Feb 16 2022 at 17:36):

Any homomorphism that's conjugate to an isomorphism is an isomorphism, so by the 2-out-of-6 property all the equivalences in this 2-category arise from some isomorphism of groups. The point is weaker than that, that such an equivalence doesn't determine a unique isomorphism.

John Baez (Feb 16 2022 at 17:48):

David Egolf said:

Thanks so much for the detailed explanation! It's going to take me a bit of time and effort to absorb it.

Heh, all this stuff shows up when you try to be very clear about what you mean by "a groupoid with two objects". Do you mean it has the property of having two objects but maybe you don't have a way to refer to either of the objects? Do you mean it has the structure of having two specified objects, say $x$ and $y$, that you know how to refer to? Etc. These subtle but crucial distinctions are made precise when we form a category of groupoids with two objects: different ways of thinking give really different categories.

And unfortunately, you chose an example where it's almost irresistible to form a 2-category, so we get even more choices of what to do!

John Baez (Feb 16 2022 at 17:49):

(But we don't have to use a 2-category.)

David Egolf (Feb 17 2022 at 16:50):

Mike Shulman said:

But things are different for one-object groupoids. A functor between one-object groupoids is the same regardless of whether the one objects are specified or not: it's just a group homomorphism between the automorphism groups. But a transformation between functors $f,g:G\to H$ involves a function ${\rm ob}(G) \to {\rm mor}(H)$ giving the components of the transformation, and even if ${\rm ob}(G)$ is a one-element set, the same is not true of ${\rm mor}(H)$. However, if $\star$ is the one object of $H$, then $H(\star)$ does have a distinguished element, namely the identity ${\rm id}_\star$. Thus, a transformation between pointed one-object groupoids must be the identity, and so the 2-category of pointed one-object groupoids is equivalent to the 1-category of groups.

I'm trying to understand this.
Let's say $G$ and $H$ are one-object groupoids.
A functor $f: G \to H$ corresponds to a map $f_o: {\rm ob}(G) \to {\rm ob}(H)$ and a map $f_m: {\rm mor}(G)\to {\rm mor}(H)$.
The map $f_o$ is forced to sends the one object of $G$ to the one object of $H$.

Let $g: G \to H$ also be a functor. A natural transformation $\alpha: f \to g$ includes as part of its data an assignment to each object in $G$ (call the singular object $\star$) a morphism $\alpha_\star$ in ${\rm mor}(H)$. So, a natural transformation $\alpha$ specifies by $\alpha_\star$ an element of ${\rm mor}(H)$.

We also note that among the morphism in ${\rm mor}(H)$, there is a distinguished morphism - namely the identity morphism. So, what - if any - relationship must there be between these two elements of ${\rm mor}(H)$, namely the identity map and $\alpha_\star$? I'm guessing there is some way to show that $\alpha_\star$ must be the identity on $H$, but I don't see it right now.
Although maybe I'm wrong, and this isn't what @Mike Shulman is describing here.

Mike Shulman (Feb 17 2022 at 16:59):

In that comment I was just saying that since, intuitively, any operation acting between pointed sets ought to be required to preserve the points, this should also be true of $\alpha$, and thus it should take the point $\star$ to the point $\rm id_\star$.

Mike Shulman (Feb 17 2022 at 17:00):

It's not an automatic statement about any transformation between such functors -- in fact, that's exactly the point -- rather, it's a condition that we should impose on such transformations if we regard the one-object groupoids as "pointed".

Mike Shulman (Feb 17 2022 at 17:01):

A more abstract way to derive this is to say that just as the category of pointed sets is the coslice $1/Set$, the 2-category of pointed groupoids should be the coslice $1/Gpd$. If you work out the 2-cells in a coslice category you'll find that this condition pops out.

David Egolf (Feb 17 2022 at 17:12):

Ok, that makes sense! Thanks!
This reminds me a little of what I was learning about algebras, where a "pointed" element corresponds to a map $1 \to S$. If an algebra is a map $f: 1+S^2 \to S$ for some set $S$, then this corresponds to a distinguished element of $S$ together with some binary operation on $S$. Then, a morphism of algebras is required to send distinguished elements to distinguished elements.
The coslice category $1/Set$ should have as objects maps $1 \to S$ for different sets $S$. Then morphisms are maps between sets that send the distinguished element of one set to another.
The coslice category $1/Gpd$ should have as objects maps from the terminal groupoid, which I think is the identity group. So, an object in this coslice category is a groupoid together with a distinguished object (and a distinguished identity morphisms on that object). A map between these objects is a functor between groupoids that must send the distinguished object to the distinguished object.
I don't really know what a 2-cell is, though. But I can imagine it involves a commutative square of pointed groupoids... and I can imagine that what we're talking about would indeed show up there!

Mike Shulman (Feb 18 2022 at 17:40):

You could check out [[slice 2-category]] if it would help...

David Egolf (Feb 18 2022 at 17:44):

Mike Shulman said:

You could check out [[slice 2-category]] if it would help...

Thanks, appreciate the link! This will be helpful once I learn basic things, like what a "2-morphism" in a "non-strict 2-category" is.

Mike Shulman (Feb 18 2022 at 18:01):

A "non-strict 2-category" just means a [[bicategory]], and a 2-morphism there is the same as a 2-morphism in a strict 2-category. The definition of slice works just fine in the strict case too.