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It is well-known that every connected groupoid is equivalent to a group. However the 2-category of connected groupoids is not equivalent to the 2-category obtained trivially from the 1-category of groups. This clashes with my intuition: the first statement is intuitively saying that a connected groupoid is the same thing as a group, and the second is intuitively saying that the collection of all connected groupoids somehow behaves different than the collection of all groups. What causes this mismatch? Is thinking of equivalent objects in math as "the same" bad, and if so, how should we think about equivalent objects?

Carla only proves trivial prop
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  • I dig your username. I also usually skip the complex proofs. :) And your avatar is something else... Commutative diagrams are beautiful. – EffingLoveMath Aug 26 '22 at 22:56
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    My initial impression would be that the natural way to try to construct an equivalence would be to choose a "base point" for each groupoid, but presumably there's no way to do so in a sufficiently consistent way to get an actual equivalence of 2-categories. – Daniel Schepler Aug 26 '22 at 22:59
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    Regarding the user name, I think Grothendieck once said (paraphrasing) that if you have complex proofs, you haven't developed the underlying theory sufficiently yet; and once you have properly developed a theory, every important result should "look" trivial to build using the previous results and the general methods. – Daniel Schepler Aug 26 '22 at 23:01
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    @DanielSchepler While indeed i think thats the problem, it still feels jarringly wrong that a collection of objects that are "the same" don't behave the same way, and so there must be something wrong with my intuition of thinking of equivalent objects as "the same". – Carla only proves trivial prop Aug 26 '22 at 23:04
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    Every connected groupoid is equivalent to a group but not in a unique way. You must choose a basepoint. The 2-category of pointed connected groupoids is equivalent to the category of groups. – Zhen Lin Aug 26 '22 at 23:14
  • Those pesky automorphisms... Once again producing an obstruction to something that seems like it should be possible... – Daniel Schepler Aug 26 '22 at 23:18

1 Answers1

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the first statement is intuitively saying that a connected groupoid is the same thing as a group

No it's not. It's saying that isomorphism classes of connected groupoids can be identified with isomorphism classes of groups. But an equivalence of categories or $2$-categories needs to consider morphisms (and $2$-morphisms), not just objects. You can show that the $2$-category of connected groupoids is equivalent to the $2$-category whose

  • objects are groups $G$,
  • morphisms are group homomorphisms $f : G \to H$, and
  • $2$-morphisms between two parallel group homomorphisms $f, g : G \to H$ are elements $h \in H$ such that $hf(x) h^{-1} = g(x)$ for all $x \in G$.

It's the $2$-morphisms here that make this $2$-category not equivalent to the category of groups. For example, $\text{Hom}(\mathbb{Z}, G)$ is the groupoid whose objects are the elements of $G$ and whose morphisms are conjugacies between these elements, and $\text{Hom}(\text{id}_G, \text{id}_G)$ (the $2$-morphisms from the identity homomorphism $\text{id}_G : G \to G$ to itself) is the center $Z(G)$.

What is true is that the $2$-category of pointed connected groupoids is equivalent to the category of groups; the addition of basepoints removes the interesting $2$-morphisms above and allows us to write down a canonical functor to $\text{Grp}$ given by taking the automorphism group of the basepoint. Without the basepoint, to actually write down a group starting from a connected groupoid you have to pick a basepoint, and there isn't a canonical one.

This distinction between connected things and pointed connected things is much more significant than it might appear at first glance and rears its head in many other ways. For example, you can define what it means for a group $G$ to act on a connected groupoid $X$, and you can find examples of such actions that don't arise from actions of $G$ on the group $\pi_1(X, x)$. This is because actions arising from an action on $\pi_1$ need to have a fixed point and they don't always; see e.g. this MO answer. (I am being a bit sloppy here; there are two possible definitions of an action and also two possible definitions of a fixed point depending on $2$-categorical you're willing to be.)

Here's another example: recall that the absolute Galois group of a field $K$ is the Galois group $\text{Gal}(\bar{K}/K)$. It behaves in many ways like a fundamental group (of the affine scheme $\text{Spec } K$), and is generalized by the etale fundamental group. It would be nice if it were a (contravariant) functor. But it isn't! The reason is that the algebraic closure is not actually unique, only unique up to (non-unique) isomorphism, and to define the absolute Galois group you have to pick an algebraic closure, which is the analogue in this setting of picking a basepoint. The notation is a bit deceptive here because we write the algebraic closure $\bar{K}$ as if it were a functor but it isn't one either.

The functorial construction is not a group but a connected groupoid, namely the connected groupoid of all algebraic closures of $K$. And now the functor is easy to write down: given a morphism $K \to L$ of fields, every algebraic closure $\bar{L}$ of $L$ canonically induces an algebraic closure of $K$, namely the algebraic closure of $K$ in $\bar{L}$.

Qiaochu Yuan
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    Here is a simpler example of the same kind of phenomenon: we know that every $n$-dimensional real vector space is isomorphic to $\mathbb{R}^n$. It certainly does not follow that the category of $n$-dimensional real vector spaces is equivalent to the category with a single object $\mathbb{R}^n$ and no non-identity morphisms, because that disregards the morphisms! Picking a basis of a vector space is analogous to picking a basepoint in the above discussion; it can be done but not canonically and usually not in a way that's compatible with extra structure. – Qiaochu Yuan Aug 26 '22 at 23:30
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    As a picture of a similar phenomenon, maybe think of the Meobius strip as a vector bundle over the circle: every fiber is isomorphic to $\mathbb{R}$, but the whole vector bundle is not isomorphic to $\mathbb{R} \times \mathbb{S}^1$. – Daniel Schepler Aug 26 '22 at 23:49
  • If we shouldn't think of equivalent objects as "the same", is there a better intuition for how to think about them? Without this intuition its not clearly to me philosophically why or when do we want to preserve equivalences. – Carla only proves trivial prop Aug 26 '22 at 23:51
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    @Carla: one shouldn't think in terms of equivalent objects but in terms of equivalences; when you say "equivalent" you record only the one bit of information of whether an equivalence exists or not, whereas in practice we want to know that specific maps are equivalences, so that we can actually use them (and also e.g. so we don't get confused and try to use two different equivalences without noticing that we've done so). This is exactly the move from thinking about isomorphism classes to thinking about categories, with their objects and their morphisms. – Qiaochu Yuan Aug 26 '22 at 23:55
  • About the first comment: isn't the category of n-dimensional real vector spaces equivalent to the category with a single object $\mathbb R^n$ with its usual linear endomorphisms? – Carla only proves trivial prop Aug 27 '22 at 00:02
  • @Carla: yes. This is called taking the skeleton; instead of having many different isomorphic objects we pick one in each isomorphism class and throw out the rest. The analogous construction for the $2$-category of connected groupoids produces the $2$-category I described above. In my first comment I ignored the morphisms and in your question you ignored the $2$-morphisms. – Qiaochu Yuan Aug 27 '22 at 00:04
  • Why do considering connected groupoids instead of groups make non-trivial 2-morphisms appear? – Carla only proves trivial prop Aug 27 '22 at 00:07
  • @Carla: this is a good exercise and I encourage you to do it. Think of groups $G$ as one-object groupoids $BG$ and write down explicitly the definition of a natural transformation between two functors $f, g : BG \to BH$. – Qiaochu Yuan Aug 27 '22 at 00:08
  • I have already done this exercise actually. But then groups also have those 2-morphisms! So why aren't groups the same as connected groupoids? – Carla only proves trivial prop Aug 27 '22 at 00:11
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    The whole point of our discussion here is exactly that groups are not the same thing as the corresponding groupoids; the category of groups does not have these $2$-morphisms, it's just the ordinary $1$-category you're familiar with. The interesting claim that has not been justified yet is why it's equivalent to the $2$-category of pointed connected groupoids; as another exercise you can try to write down what a functor between pointed connected groupoids is, then what a natural transformation of such things is, to see that there aren't any non-identity ones anymore. – Qiaochu Yuan Aug 27 '22 at 00:20
  • Does that mean that in reality groups are not the same thing as a category with 1 object where every morphism is an automorphism? Because groups don't have 2-morphisms? And if we defined groups in that way then they'd be equivalent to connected groupoids too right? If the answer to these questions is affirmative then everything would make sense philosophically to me! – Carla only proves trivial prop Aug 27 '22 at 00:23
  • Yes, because the category of groups doesn't have $2$-morphisms. For your second question it depends on whether you make the categories pointed or not (in the sense of distinguishing their single object). But again to really understand what's going on you should do this second exercise. – Qiaochu Yuan Aug 27 '22 at 00:27
  • Thank you, i will. To me a pointed category is not the same thing as a category, so i guess when people say "a group is a category with 1 object where all automorphisms are invertible" , they mean a pointed such category, right! (saying a category has 1 object wouldn't respect equivalence so it wouldn't be a good thing to say anyway, but it makes sense if its a pointed category) – Carla only proves trivial prop Aug 27 '22 at 00:36
  • Yes, strictly speaking people should say "pointed category." It doesn't make any difference on the level of isomorphism classes because there is a unique pointing, but as we've been discussing it makes a difference on the level of $2$-morphisms. – Qiaochu Yuan Aug 27 '22 at 04:11