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Stream: learning: questions

Topic: What does the associator of a special 2-group tell you?

Joshua Meyers (Apr 05 2024 at 18:42):

Let $G$ be a skeletal 2-group with strict inverses. Then by Eckmann-Hilton $\text{Aut}(1)$ is abelian. Since the map $-\otimes 1_x : \text{Aut}(1)\to\text{Aut}(x)$ is an isomorphism, all automorphism groups $\text{Aut}(x)$ are isomorphic, and hence abelian. Since $G$ is skeletal, all morphisms of $G$ are automorphisms and thus elements of these automorphism groups.

Given arbitrary morphisms $f:x\to x,\ g:y\to y,\ h:z\to z$, the naturality of the associator tells us that $(f\otimes g)\otimes h + \alpha_{x,y,z} = \alpha_{x,y,z} + f\otimes (g\otimes h)$, in other words $(f\otimes g)\otimes h = f\otimes (g\otimes h)$ (since all automorphism groups are abelian). So tensor is strictly associative on morphisms just as it is strictly associative on objects.

The above considerations lead me to the conclusion that the associator "stands apart" from the rest of the structure of the skeletal 2-group as "extra garbage" [cf. the extra garbage discussed here.

What am I missing? In what way does the associator "matter"?

Joshua Meyers (Apr 05 2024 at 18:49):

Perhaps the associator only really starts to matter once we consider non-special 2-groups in the equivalence class of $G$?

Joshua Meyers (Apr 05 2024 at 18:51):

But if "equivalence" is the right notion of sameness for 2-groups, we should be able to see all the relevant structure just looking at special 2-groups and in a special 2-group all the actual operations $\otimes$, $\circ$, $1$, etc. seem to take place without involving the associator at all!

Mike Shulman (Apr 05 2024 at 19:01):

It's a cohomology class!

Joshua Meyers (Apr 05 2024 at 19:01):

I know but what does it tell you?

Mike Shulman (Apr 05 2024 at 19:02):

I'm not sure what kind of answer you want.

Mike Shulman (Apr 05 2024 at 19:04):

Since you said "I know" I guess you are familiar with the classification of 2-groups in terms of the group $G_0$ of objects, the abelian group $\mathrm{Aut}(1)$, the action of $G_0$ on $\mathrm{Aut}(1)$ by conjugation, and the class in group cohomology $H^3(G_0, \mathrm{Aut}(1))$ determined by the associator?

Joshua Meyers (Apr 05 2024 at 19:04):

Yes I am familiar with that classification

Joshua Meyers (Apr 05 2024 at 19:05):

That classification is my first exposure to cohomology which might shed light on my confusion

Mike Shulman (Apr 05 2024 at 19:05):

From that perspective, it's part of the data that determines the 2-group. What else do you want it to tell you?

Joshua Meyers (Apr 05 2024 at 19:05):

I am surprised that it doesn't make any computation in the 2-group happen differently, other than in a tautological sense (the computation "compute the associator")

Joshua Meyers (Apr 05 2024 at 19:06):

I would think 2-groups with different associators would be different in some meaningful way

Mike Shulman (Apr 05 2024 at 19:07):

I wouldn't say that. There are plenty of real calculations that can be done in a 2-group that involve the associator. It's true that in the skeletal case you could do the same calculation by omitting the associators, but that wouldn't be doing the calculation in the 2-group you had, but rather a different one with a different associator (which happens to be the identity in the skeletal case).

Joshua Meyers (Apr 05 2024 at 19:08):

How can I witness the (non-tautological) ramifications of choosing a different associator on the structure of a 2-group?

Mike Shulman (Apr 05 2024 at 19:11):

Pick anything you can do in a monoidal category involving three or more objects, it will probably involve the associator.

Joshua Meyers (Apr 05 2024 at 19:14):

well if we just consider skeletal 2-groups, it seems like you can do everything you want to even involving three or more objects without using the associator...

Mike Shulman (Apr 05 2024 at 19:16):

My point is that any kind of "thing you can do in a monoidal category" must be phrased in a way that involves the associator. When you specialize such a thing to a monoidal category that happens to be skeletal, the associator is still there; it doesn't magically disappear because you could have performed the same construction in a different monoidal category that happens to have the same underlying category.

Joshua Meyers (Apr 05 2024 at 19:16):

I guess my question is, if we hold the triple $(G_0, \text{Aut}(1), a:G_0\to\text{Aut}(\text{Aut}(1)))$ constant, what are the stakes of choosing one associator rather than another?

Joshua Meyers (Apr 05 2024 at 19:17):

Your responses make me think that to understand the significance of the associator, I need to think about possibly non-skeletal 2-groups

Joshua Meyers (Apr 05 2024 at 19:19):

But that also strikes me as strange, as if (1) we are using the right notion of equivalence of 2-groups, and (2) every 2-group is equivalent to a special 2-group, then we should be able to understand 2-groups in general by thinking only about special 2-groups

Mike Shulman (Apr 05 2024 at 19:21):

It's certainly possible, but I think this may be a case of where working in more generality is actually easier, because accidental coincidences in a special case can make some things typecheck that really shouldn't.

John Baez (Apr 05 2024 at 21:07):

Joshua Meyers said:

What am I missing? In what way does the associator "matter"?

In mych the same sort of way that the multiplication in a group matters. For example, groups correspond to connected pointed homotopy 1-types (very roughly, connected pointed topological spaces whose homotopy groups vanish above the first homotopy group). Similarly, 2-groups correspond to connected pointed homotopy 2-types. So, two special 2-groups that are the same except for their associator will give different-looking spaces.

Joshua Meyers (Apr 05 2024 at 21:15):

How do I contruct the space corresponding to a special 2-group?

John Baez (Apr 05 2024 at 21:19):

First read my papers...

Joshua Meyers (Apr 05 2024 at 21:20):

I've read (most of) these papers:

• Higher-Dimensional Algebra V: 2-Groups, by John Baez

• Hoàng Xuân Sính’s Thesis: Categorifying Group Theory, by John Baez

• Lectures on n-Categories and Cohomology, talks by John Baez, notes by Michael Shulman

Are they the ones you mean or did I miss one?

Joshua Meyers (Apr 05 2024 at 21:24):

You could also point me to a particular section of one of these papers, maybe I skipped it or didn't understand when I read it...

Mike Shulman (Apr 05 2024 at 21:31):

One way to construct the space corresponding to a 2-group $G$ is as follows. Define it as a simplicial set and then take the geometric realization. In the simplicial set:

1. There is one 0-simplex.
2. The 1-simplices are objects of $G$.
3. The 2-simplices with boundary $x,y,z$ are morphisms $x\otimes z \to y$.
4. The 3-simplices with boundary $f_{012}:x_{01} \otimes x_{12} \to x_{02}$, $f_{013} :x_{01}\otimes x_{13} \to x_{03}$, $f_{023} : x_{02} \otimes x_{23} \to x_{03}$, and $f_{123} : x_{12} \otimes x_{23} \to x_{13}$ are equalities between $f_{023} \circ (f_{012} \otimes 1_{x_{23}}) = f_{013} \circ (1_{x_{01}} \otimes f_{123}) \circ \alpha$, where $\alpha$ is the associator.
5. The higher simplices are all trivial.

So you can see how the associator gets used.

John Baez (Apr 05 2024 at 21:40):

Joshua Meyers said:

I've read (most of) these papers:

• Higher-Dimensional Algebra V: 2-Groups, by John Baez

• Hoàng Xuân Sính’s Thesis: Categorifying Group Theory, by John Baez

• Lectures on n-Categories and Cohomology, talks by John Baez, notes by Michael Shulman

Are they the ones you mean or did I miss one?

Yes. The paper about Sinh's thesis gives the clearest explanation of how to go back and forth between pointed spaces and 2-groups, I think. Near the end (but before the appendix.)

John Baez (Apr 05 2024 at 21:48):

Mike just explained one direction. But that paper does the explanation with pictures, and I think they help.

John Baez (Apr 05 2024 at 21:50):

You'll see how group elements give edges, multiplication gives triangles, 2-group morphisms and associators give tetrahedra, etc.

Joshua Meyers (Apr 05 2024 at 21:51):

I see. So now it seems I need to understand Postnikov invariants

Joshua Meyers (Apr 05 2024 at 21:54):

It would be great if there was a really tiny example where I could try out different associators and actually visualize the corresponding spaces.

John Baez (Apr 05 2024 at 21:59):

Joshua Meyers said:

I see. So now it seems I need to understand Postnikov invariants

You are learning Postnikov invariants right now. The way the associator of a 2-group affects the corresponding space: that's the simplest example of a Postnikov invariant.

John Baez (Apr 05 2024 at 22:01):

Then comes the pentagonator of a 3-group, etc.

John Baez (Apr 05 2024 at 22:06):

Joshua Meyers said:

It would be great if there was a really tiny example where I could try out different associators and actually visualize the corresponding spaces.

For that I suggest taking $\pi_1 = G$ and $\pi_2 =H$ to be the smallest nontrivial group discovered by humanity so far. Then there are two really different choices of associator, giving two inequivalent 2-groups.

Joshua Meyers (Apr 06 2024 at 14:55):

Here is what I got for $\pi_1$ and $\pi_2$ both the cyclic group $\{0,1\}$ of order 2 and the trivial action (it is indeed the only action). There are two possible associators, giving two inequivalent 2-groups, giving two inequivalent spaces.
2024-04-06-105352.jpg

Joshua Meyers (Apr 06 2024 at 14:56):

In this picture, $\wedge$ is "and", defined by $x\wedge y = 1$ iff $x=1$ and $y=1$, btw

Joshua Meyers (Apr 06 2024 at 14:59):

Now, what I don't know how to do, even in this simple example, is visualize the spaces enough to see what difference the circled $+1$ makes. When I try, I start to think of each $0,1$ pair of 2-cells as a sphere with three points pulled out of it and joined together. But then I have no idea how to think of all eight 2-cells together, and that doesn't even get started on the tetrahedra. Is there a topology text I should study that would make me able to visualize spaces like this?

Joshua Meyers (Apr 06 2024 at 15:29):

Or another possible answer is that people don't usually visualize these spaces, they just leave them as descriptions on the paper. But I would hope that in as simple a case as this, the spaces would be susceptible to some nicer description which I could begin to visualize.

John Baez (Apr 06 2024 at 17:24):

People don't usually visualize these spaces except in a very subtle sense of "visualize", because they are infinite-dimensional.

John Baez (Apr 06 2024 at 17:25):

There are things you can do to better understand them, though.

John Baez (Apr 06 2024 at 17:27):

I'd start by pondering the famous 'Eilenberg-Mac Lane spaces' $K(\mathbb{Z}/2,1)$ and $K(\mathbb{Z}/2,2)$.

• $K(\mathbb{Z}/2,1)$ is the connected pointed space whose only nonvanishing homotopy group is $\pi_1 = \mathbb{Z}/2$.

• $K(\mathbb{Z}/2,2)$ is the connected pointed space whose only nonvanishing homotopy group is $\pi_2 = \mathbb{Z}/2$.

Both the spaces you're talking about, say $X$ and $X'$ are built from $K(\mathbb{Z}/2,1)$ and $K(\mathbb{Z}/2,2)$.

In technical terms, we say both $X$ and $X'$ aretotal spaces of bundles over $K(\mathbb{Z}/2,1)$, and both have the same fiber, which is called $K(\mathbb{Z}/2,2)$.

John Baez (Apr 06 2024 at 17:28):

In other words we have a map

$p: X \to K(\mathbb{Z}/2,1)$

and this map $p$ looks locally like the projection from $K(\mathbb{Z}/2,1) \times K(\mathbb{Z}/2,2)$ to $K(\mathbb{Z}/2, 1)$. And the same is true for $X'$!

John Baez (Apr 06 2024 at 17:31):

So your spaces $X$ and $X'$ are very similar; the only difference is that one of them - the one with the trivial associator - actually is the product $K(\mathbb{Z}/2,1) \times K(\mathbb{Z}/2,2)$, while the other - the one with the nontrivial associator - is only locally a product.

John Baez (Apr 06 2024 at 17:38):

(What I'm doing now is teaching you the theory of 'Postnikov towers' in an example, namely the example where $\pi_1$ and $\pi_2$ happen to be $\mathbb{Z}/2$. Everything I'm saying works more generally.)

John Baez (Apr 06 2024 at 17:40):

Now $K(\mathbb{Z}/2,1)$ is an extremely important space, and there are lots of ways to visualize it and understand it. For one thing, we can show that this space is $\mathbb{R}\mathrm{P}^\infty$, the infinite-dimensional real projective space.

John Baez (Apr 06 2024 at 17:41):

One way to understand this is to build it up using simplices and show that the $n$-dimensional approximation to this space, where you go up to the $n$-simplices, is $\mathbb{R}\mathrm{P}^n$, the n-dimensional real projective space.

John Baez (Apr 06 2024 at 17:42):

This is tons of fun, but probably more fun if you've already thought a bit about real projective space.

John Baez (Apr 06 2024 at 17:46):

One reason this stuff is tons of fun is that it involves an idea you've already invented: using logical operations on bits! Given what you've already done, it should not come as an utter and complete shock that we can describe $K(\mathbb{Z}/2,1) \simeq \mathbb{R}\mathrm{P}^\infty$ using bit strings!

Joshua Meyers (Apr 12 2024 at 15:48):

I will get to this in about a week and a half! I didn't forget about it, just preoccupied with other things

Joshua Meyers (Apr 23 2024 at 01:19):

I will work out the topology soon, but I just wanted to say I found another way of thinking about this, through strictifications. Let $N$ be a strict monoidal category whose objects are the monoid $\mathbb{N}$ which is monoidally equivalent to our skeletal 2-group $G$. Then $N$ is a groupoid and each object has exactly 2 endomorphisms. There are two connected components, even numbers and odd numbers. Let $a:0\to 0$ be non-identity, and let $x:0\to 2$ and $y:2\to 0$ be chosen to correspond to the identity on the unit of $G$ under the equivalence.

Then the morphisms of $G$ are generated by $a,x,y$. This is easier to show with string diagrams, which can be used without guilt since $G$ is a PRO. We denote $a$ as a dot, $x$ as a cap, and $y$ as a cup. Because of the equivalence, we must have $x$ and $y$ inverse, so in the string diagrams, a circle can vanish, and a pair of vertical parallel lines can split into a cup and a cap. We then note that a dot is equal to a double-circled dot, so a circled dot must also be non-identity, and thus must be equal to a dot. We are then confronted with the question of the zigzag -- when you pull it straight to a vertical line, does a dot pop out beside it or not? It can be shown that this choice completely determines the rest of the structure of $N$, and both choices are consistent.

I am confident that the two choices of the zigzag correspond to the two choices of the associator, but I have not proven the existence of these equivalences.

Joshua Meyers (Apr 23 2024 at 01:44):

Update: I did prove their existence

John Baez (Apr 23 2024 at 15:53):

Cool. I will have to think about this.