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Stream: learning: questions

Topic: free monoidal category on one generator

Matteo Capucci (he/him) (Sep 09 2021 at 09:41):

It's easy to see that the free monoid on one generator is $\mathbb N$ itself, and incidentally it is commutative since $1$ is the only generator and of course $1+1=1+1$ .
What is the free monoidal category on one generator (call it $X$) then? It should be easily constructed: it's the category whose objects are parenthesized strings of $X$ s (call the empty string $I$) and whose morphisms are all given by $\alpha$, $\lambda$ and $\rho$, which are formally adjoined (together with the pentagonal and triangular relations).
I believe this category, call it $\mathscr N$, should have the following properties:

• First, it is actually a preorder, since by MacLane's theorem, all diagrams made from $\alpha, \lambda, \rho$ commute, hence any two parallel arrows are equal
• Second, it is a groupoid, since all the generating morphisms are invertible
• Third, it is symmetric, with braiding given by $X \otimes X = X \otimes X$ and $X \otimes I \xrightarrow{\rho} X \xrightarrow{\lambda} I \otimes X$ (so that's why it is not commutative). The hexagon identities are trivially satisfied bc the category is a preorder.

So let's try to contract the isomorphism classes of this category to obtain a more succint presentation. We get a category whose objects are natural numbers (there are no morphisms from words of different 'length' [ignoring the $I$ s] in $\mathscr N$), and whose morphisms are reparenthesizations (I apologise for the spelling). This feels like it ought to be the permutation groupoid, aka the core of FinSet. Indeed, on its nlab page, it's stated that $\mathbb P$ is the 'free strict symmetric monoidal category'.

But how does a reparenthesization give rise to a permutation?
And, most importantly, is it there a more succint/conceptual presentation of $\mathscr N$ itself, i.e. which doesn't strictify the monoidal structure?

Chad Nester (Sep 09 2021 at 09:47):

Abramsky's has a paper that deals with these questions in some detail.

Chad Nester (Sep 09 2021 at 09:48):

The free monoidal and symmetric monoidal category on $1$ are singled out, since they highlight the difference between the two constructions. In particular the free monoidal category is as you describe it, but the free symmetric monoidal category is the permutation groupoid.

Jules Hedges (Sep 09 2021 at 09:52):

(nb. If you didn't care about strictness then you missed the extremely easy way to deal with free strict monoidal categories, namely that it's a very degenerate category of string diagrams. You can directly visualise its properties from there)

Jules Hedges (Sep 09 2021 at 09:53):

I'm quite skeptical of your claim that the free monoidal category is symmetric (ignoring the minor thing that it doesn't typecheck!)

James Wood (Sep 09 2021 at 10:05):

The free monoidal category on the discrete 1-object category

Matteo Capucci (he/him) (Sep 09 2021 at 10:38):

Thanks for the ref Chad!
Chad Nester said:

The free monoidal and symmetric monoidal category on $1$ are singled out, since they highlight the difference between the two constructions. In particular the free monoidal category is as you describe it, but the free symmetric monoidal category is the permutation groupoid.

The free monoidal category on one object, Abramsky claims, is the discrete monoidal category $(\mathbb N, =, 0, +)$. Is it what I described? :thinking: (EDIT: I see, he considers strict monoidal structures, but then this doesn't answer my question!)

Matteo Capucci (he/him) (Sep 09 2021 at 10:38):

Jules Hedges said:

I'm quite skeptical of your claim that the free monoidal category is symmetric (ignoring the minor thing that it doesn't typecheck!)

Can you elaborate? What is it that doesn't typecheck?

Zhen Lin Low (Sep 09 2021 at 10:43):

Being symmetric is not a property but rather a structure, so you are supposed to also give the symmetry isomorphism, I guess.

Zhen Lin Low (Sep 09 2021 at 10:45):

I think Mac Lane's coherence theorem implies that the free monoidal category on 1 is equivalent to the free strict monoidal category on 1, which obviously admits the structure of (strict!) symmetric monoidal category in a unique way.

James Wood (Sep 09 2021 at 10:51):

Matteo Capucci (he/him) said:

The free monoidal category on one object, Abramsky claims, is the discrete monoidal category $(\mathbb N, =, 0, +)$. Is it what I described?

The core of FinSet is the free symmetric monoidal category on one object, which, as noted, is different to the free monoidal category on one object. The core of order-preserving FinSet would be one way to describe the free monoidal category on one object, and this is just $(\mathbb N, =, 0, +)$. Equivalence of categories preserves the fact that the category is a poset, and the core of FinSet is not.

Jules Hedges (Sep 09 2021 at 11:35):

Zhen Lin Low said:

Being symmetric is not a property but rather a structure, so you are supposed to also give the symmetry isomorphism, I guess.

Right, you could say that the free monoidal category can be equipped with a symmetry (which is the main thing I'm doubting), but it's technically sketchy to say that a category equipped with a choice of symmetry satisfies the universal property of a free monoidal category without symmetry

Jules Hedges (Sep 09 2021 at 11:35):

An example of this is in Abramsky's paper (going from memory here, this might not be exactly right): the free monoidal category on a category can be canonically equipped with a trace, but it's not the free traced monoidal category - that also exists but has a different underlying category

Jules Hedges (Sep 09 2021 at 11:50):

(For example, the free monoidal category on a category always has exactly one morphism $I \to I$, but the free traced monoidal category generally has many non-identity scalars)

Zhen Lin Low (Sep 09 2021 at 11:51):

Sure, of course. But I think that goes without saying.

Jules Hedges (Sep 09 2021 at 11:57):

Zhen Lin Low said:

Sure, of course. But I think that goes without saying.

On #learning: questions things that go without saying are often worth saying

John Baez (Sep 09 2021 at 15:12):

Matteo Capucci (he/him) said:

It's easy to see that the free monoid on one generator is $\mathbb N$ itself, and incidentally it is commutative since $1$ is the only generator and of course $1+1=1+1$ .
What is the free monoidal category on one generator (call it $X$) then? It should be easily constructed: it's the category whose objects are parenthesized strings of $X$ s (call the empty string $I$) and whose morphisms are all given by $\alpha$, $\lambda$ and $\rho$, which are formally adjoined (together with the pentagonal and triangular relations).
I believe this category, call it $\mathscr N$, should have the following properties:

• First, it is actually a preorder, since by MacLane's theorem, all diagrams made from $\alpha, \lambda, \rho$ commute, hence any two parallel arrows are equal
• Second, it is a groupoid, since all the generating morphisms are invertible.
• Third, it is symmetric, with braiding given by $X \otimes X = X \otimes X$ and $X \otimes I \xrightarrow{\rho} X \xrightarrow{\lambda} I \otimes X$ (so that's why it is not commutative). The hexagon identities are trivially satisfied bc the category is a preorder.

This sounds correct.

So let's try to contract the isomorphism classes of this category to obtain a more succinct presentation.

I'd prefer to say either "take a skeleton" or "strictify", since those are two things you can always do to a monoidal category. The latter gives a strict monoidal category equivalent to the monoidal category you started with.

However, in the particular case you're looking at, a monoidal category where each component is a contractible groupoid, taking a skeleton and strictifying give isomorphic strict monoidal categories - and yes, you can also describe this process as "contracting the isomorphism classes" to get an equivalent discrete category!

So, the particular monoidal category you mentioned is equivalent to the strict monoidal category with natural numbers as objects, only identity morphisms, addition as tensor product, 0 as the unit object, and (necessarily) identities as associator and unitors.

In other words: the free monoidal category on 1 is equivalent, as a monoidal category, to the free strict monoidal category on 1.

We get a category whose objects are natural numbers (there are no morphisms from words of different 'length' [ignoring the $I$ s] in $\mathscr N$), and whose morphisms are reparenthesizations (I apologise for the spelling). This feels like it ought to be the permutation groupoid, aka the core of FinSet. Indeed, on its nlab page, it's stated that $\mathbb P$ is the 'free strict symmetric monoidal category.

It doesn't feel like that to me! The free strict symmetric monoidal category on 1 is the permutation groupoid, aka the core of FinSet. But you're talking about the free monoidal category on 1, and that's a different thing.

It's a kind of freak of low dimensions that the free monoid on 1 is the free commutative monoid on 1. When we categorify, and switch from the property of commutativity to the structure of a symmetry, we no longer get this coincidence: the free symmetric monoidal category on 1 is different from the free monoidal category on 1. And there's another interesting thing, intemediate between these two and different from both: the free braided monoidal category on 1. This is the braid groupoid.

Jules Hedges (Sep 09 2021 at 15:20):

And to expand on what I said early about free strict monoidal categories being easy to understand: The free strict monoidal category on a single object generator is (by the Joyal-Street theorem) the category whose morphisms are progressive, planar string diagrams with no nodes. From there you can directly see that its objects are basically natural numbers (counting how many strings there are), and it only has identity morphisms. There's no way to construct a progressive string diagram $m \to n$ for $m \neq n$ if you don't have any nodes. And the only string diagram $n \to n$ is just $n$ parallel strings, which is the identity morphism

Jules Hedges (Sep 09 2021 at 15:27):

(Here "progressive" is Joyal-Street's technical term meaning "no cups or caps", strings can only go forwards)

Matteo Capucci (he/him) (Sep 10 2021 at 08:00):

Thanks @John Baez and @Jules Hedges, I see why I was struggling with this question: In didn't consider symmetry a structure but a property

Matteo Capucci (he/him) (Sep 10 2021 at 08:00):

I should have avoided this mistake

John Baez (Sep 10 2021 at 13:30):

The yoga of properties, structure and stuff is endlessly fascinating....

John Baez (Sep 10 2021 at 13:33):

For a monoid to be commutative is just a property.

For a monoidal category to be braided is structure - but then for the braided monoidal category to be symmetric is just a property.

We are zig-zagging down the periodic table here:

Nathanael Arkor (Sep 10 2021 at 13:33):

I'd be interested to know of some examples of monoidal categories with two distinct symmetries. Usually there's an "obvious" symmetry, which perhaps lends to the misconception that a symmetry is a property.

John Baez (Sep 10 2021 at 13:36):

A super-important example is the category of G-graded vector spaces for a group G. There's a sort of obvious tensor product of objects here, but then there are different ways to make this category into a monoidal category, or a braided monoidal category, or a symmetric monoidal category... and these are classified using the cohomology of G.

Nathanael Arkor (Sep 10 2021 at 13:36):

(One example is given here, for graded vector spaces.)

Nathanael Arkor (Sep 10 2021 at 13:37):

Ah, John just beat me to it!

John Baez (Sep 10 2021 at 13:38):

Yes. The math behind this example was explained in Joyal and Street's original paper Braided monoidal categories, which has more in it than their later paper Braided tensor categories.

John Baez (Sep 10 2021 at 13:38):

These days this example has become important in condensed matter physics.

John Baez (Sep 10 2021 at 13:38):

The easiest case is G $= \mathbb{Z}/2$.

John Baez (Sep 10 2021 at 13:39):

So, you look at vector spaces that are split into two parts, called in physics a "bosonic" and "fermionic" part.

John Baez (Sep 10 2021 at 13:39):

The usual tensor product of vector spaces generalizes to $\mathbb{Z}/2$-graded vector spaces.

John Baez (Sep 10 2021 at 13:40):

But then there are two choices of symmetry: the usual symmetry, or the symmetry where you stick in an extra minus sign when you switch two "fermionic" vectors.

John Baez (Sep 10 2021 at 13:41):

Nature has chosen the second symmetry, and this is why chemistry exists: with the first, all the electrons in an atom would sit in the lowest shell.

(With the second, when you switch two identical fermionic vectors you get minus that same thing, so it winds up being impossible to have two electrons in the same state, so they wind up needing to occupy different shells, giving different elements very different chemical properties.)

John Baez (Sep 10 2021 at 13:45):

In modern condensed matter physics they are exploring groups more complicated than $\mathbb{Z}/2$, where there can be braidings that aren't symmetries... and the different choices of braiding correspond to strange new kinds of matter where instead of just bosons and fermions there are "anyons" that behave in more interesting ways when you switch two of them.