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Stream: theory: category theory

Topic: the compact closed category of integers

Chad Nester (Sep 29 2020 at 13:00):

The free compact closed category on a traced monoidal category $\mathbb{X}$ is often written $\mathsf{Int}(\mathbb{X})$, A notation established in the 1996 paper introducing traced monoidal categories. While it isn't really discussed in the paper, the reason is surely that by starting with a very simple traced monoidal category $\mathbb{N}$ whose objects are natural numbers, with $\otimes = +$ and $I = 0$, we obtain a similar category of integers as $\mathsf{Int}(\mathbb{N})$. Does this appear explicitly anywhere? I'm writing a computer science paper and I want to cite something for $\mathsf{Int}(\mathbb{N})$.

Exequiel Rivas (Sep 29 2020 at 13:11):

There was some related discussion in the TYPES mailing list, under the topic "What algebra am I thinking of?", maybe it helps:
http://lists.seas.upenn.edu/pipermail/types-list/2018/002041.html
http://lists.seas.upenn.edu/pipermail/types-list/2018/002034.html

Chad Nester (Sep 29 2020 at 13:17):

That is indeed helpful -- thanks!

Jules Hedges (Sep 29 2020 at 13:25):

I was always a bit surprised it's not discussed in the original paper, since it's obviously in the background

Chad Nester (Sep 29 2020 at 18:48):

While we’re here, something related: Abramsky has a paper (2005) where he observes that the free symmetric monoidal category $\mathsf{FSM}(\mathbb{C})$ on some $\mathbb{C}$ is always traced. If we start with the terminal category then $\mathsf{FSM}(\mathbb{C})$ is a good $\mathbb{N}$ in the above sense, and the Int construction gives $\mathbb{Z}$. So we have $\mathbb{Z}$ as $\mathsf{Int}(\mathsf{FSM}(\mathbb{1}))$.

Chad Nester (Sep 29 2020 at 18:50):

So really we can get a category of integers over any category at all, and the simplest one results in the group of differences construction of the integers.

Chad Nester (Sep 29 2020 at 18:53):

:shrug:

Jules Hedges (Sep 29 2020 at 19:01):

Chad Nester said:

Abramsky has a paper (2005) where he observes that the free symmetric monoidal category $\mathsf{FSM}(\mathbb{C})$ on some $\mathbb{C}$ is always traced

This sounds extremely surprising. What sort of thing is $\mathbb C$, a set? What's the title of the paper?

Chad Nester (Sep 29 2020 at 19:03):

http://www.cs.ox.ac.uk/people/samson.abramsky/calco05.pdf

Chad Nester (Sep 29 2020 at 19:03):

$\mathbb{C}$ is any category. I also found this surprising!

Chad Nester (Sep 29 2020 at 19:05):

A fun observation is that in many ways $\mathsf{Int}(\mathsf{FSM}(1))$ captures double-entry bookkeeping. David Ellerman has a paper explaining how it’s all really about the group of differences, and the morphisms give you the permissible operations. This suggests a sort of $\mathbb{C}$-valued double entry bookkeeping...

Chad Nester (Sep 29 2020 at 19:08):

... It does come across as a bit of an “okay, so what?” thing haha.

Martti Karvonen (Sep 29 2020 at 19:29):

the thing to note (also noted in the paper) is that while the free SMC on a category $\mathbb{C}$ is canonically traced, it is not the free traced category on $\mathbb{C}$

JS PL (he/him) (Sep 30 2020 at 06:25):

Martti Karvonen said:

the thing to note (also noted in the paper) is that while the free SMC on a category $\mathbb{C}$ is canonically traced, it is not the free traced category on $\mathbb{C}$

Oops I just repeated what Martti said. But what he said is very important!

Jules Hedges (Sep 30 2020 at 09:04):

Martti Karvonen said:

the thing to note (also noted in the paper) is that while the free SMC on a category $\mathbb{C}$ is canonically traced, it is not the free traced category on $\mathbb{C}$

Huhhhh. I know this paper (I read it years ago and I cite it sometimes) but I didn't know this result, I should learn to read more carefully.....

Jules Hedges (Sep 30 2020 at 13:19):

Chad Nester said:

Abramsky has a paper (2005) where he observes that the free symmetric monoidal category $\mathsf{FSM}(\mathbb{C})$ on some $\mathbb{C}$ is always traced

I was thinking about this and I think I see morally why it's true. The morphisms of the free monoidal category on a category $\mathbb C$ should be string diagrams whose wires are labelled by objects of $\mathbb C$ and boxes are labelled by morphisms of $\mathbb C$, modulo planar isotopy + the equations of $\mathbb C$. In particular, every box has exactly one input and one output wire. That means that every use of a trace is 1 of 2 cases: either it does not create a loop and so the string can be pulled straight, or it does create a loop but in a disconnected part of the diagram

Jules Hedges (Sep 30 2020 at 13:21):

So you can define a trace by pulling the string straight in case 1, and contracting/ignoring the loop in case 2

John Baez (Sep 30 2020 at 21:31):

Nice. For us string diagram lovers that's enough of a hint to do the whole proof. I hadn't known this fact!

Chad Nester (Jul 27 2021 at 10:01):

The fact discussed above has a consequence I find surprising: If $\mathbb{C}$ is symmetric monoidal then $\mathbb{C}$ is equivalent to $\mathsf{FSM}(\mathbb{C})$. Thus every symmetric monoidal category is equivalent to a traced one.

Ivan Di Liberti (Jul 27 2021 at 10:11):

Chad Nester said:

The fact discussed above has a consequence I find surprising: If $\mathbb{C}$ is symmetric monoidal then $\mathbb{C}$ is equivalent to $\mathsf{FSM}(\mathbb{C})$. Thus every symmetric monoidal category is equivalent to a traced one.

Is there a way to see this canonical tracing in $C$? Maybe it is a trivial construction, but I do not see it.

Nathanael Arkor (Jul 27 2021 at 10:17):

Chad Nester said:

The fact discussed above has a consequence I find surprising: If $\mathbb{C}$ is symmetric monoidal then $\mathbb{C}$ is equivalent to $\mathsf{FSM}(\mathbb{C})$. Thus every symmetric monoidal category is equivalent to a traced one.

$1$ is symmetric monoidal, but $\mathsf{FSM}(1)$ is the category of finite sets and bijections.

Chad Nester (Jul 27 2021 at 10:34):

Indeed: Over lunch @Amar Hadzihasanovic convinced me that this is wrong. In particular that the free symmetric monoidal category on a symmetric monoidal category is not in general equivalent to it, as in @Nathanael Arkor 's example above.

Chad Nester (Jul 27 2021 at 10:35):

:slight_frown:

John Baez (Jul 28 2021 at 20:15):

Yeah, you shouldn't expect the free symmetric monoidal category on a symmetric monoidal category to be equivalent to it any more than you expect the free commutative monoid on a commutative monoid to be isomorphic to it.

John Baez (Jul 28 2021 at 20:15):

When you freely throw in tensor products, etc., you get a lot more stuff.