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Stream: learning: questions

Topic: Commutative Monoids and Functors from Finite Spans

Jade Master (May 27 2025 at 10:39):

In a seminar advertisement, I recently found the following claim:

Commutative monoids in a category with finite products can be
identified with product-preserving functors from spans of finite
sets.

Has anyone heard this or know the detail of how this works? Looks super interesting!

John Baez (May 27 2025 at 11:10):

The idea must be that the symmetric monoidal category of spans of finite sets is generated by spans from 2 to 1 (corresponding to multiplication in your monoid) and from 1 to 2 (corresponding to the diagonal map from your monoid to its square).

John Baez (May 27 2025 at 11:10):

Btw, I sent you an urgent email over the weekend.

Ralph Sarkis (May 27 2025 at 11:18):

You can get all the details in this paper on deconstructing Lawvere theories. It is very concisely stated in Example 6.2a, but to digest the story a bit:

Generate a category $\mathsf{F}$ of string diagrams with generators $2 \to 1$ and $0 \to 1$ and the evident equations for commutative monoids. This is the prop of commutative monoids in the sense that symmetric monoidal functor $\mathsf{F} \rightarrow \mathbf{C}$ correspond to commutative monoids in $\mathbf{C}$ .
You can also recognize that $\mathsf{F}$ is the opposite of the category of finite sets and functions.
A main theorem in that paper is that if you add generators $1 \to 2$ and $1 \to 0$ with equations for cocommutative comonoids plus a distributive law between the monoid and comonoid structures (i.e. you add a copy and delete map), you get the Lawvere theory of commutative monoids. (Theorem 6.1)
Another result in that paper is that this Lawvere theory is obtained with a distributive law between props, and there you can see that the category obtained is the category of spans of finite sets. (Example 3.4b)

You can recognize John's comments in this.

Nathanael Arkor (May 27 2025 at 11:21):

Theorem 1.24 of Gambino and Kock's Polynomial functors and polynomial monads is also worth taking a look at.

John Baez (May 27 2025 at 11:22):

I'm more familiar with the prop for commutative monoids than what we're talking about here, the Lawvere theory for commutative monoids. The prop is just FinSet $^{\text{op}}$ , since it lacks the map from 1 to 2.

James Deikun (May 27 2025 at 11:30):

Claudio Pisani (Jul 12 2025 at 09:20):

If $R$ is a rig, the Lawvere theory for $R$ -modules is the category of matrices with entries in $R$ .

Thus, in particular, the Lawvere theory for commutative monoids (which are the $\rm N$ -modules) is the category of matrices of natural numbers.

The category of spans in finite sets may be considered an unbiased version of $\rm Mat(N)$ .

John Baez (Jul 12 2025 at 09:33):

If $R$ is a rig, the Lawvere theory for $R$ -modules is the category of matrices with entries in $R$ .

Right. This can be seen from that therem saying the Lawvere theory of X's is the opposite of the category of finitely generated free X's. The Lawvere theory of $R$ -modules is thus the opposite of teh category of finitely generated free $R$ -modules, which is the category of matrices with entries in $R$ . But this category is equivalent to its opposite, by taking transposes of matrices!

John Baez (Jul 12 2025 at 09:34):

Now that makes me wonder: which Lawvere theories are equivalent to their opposite category?

Claudio Pisani (Jul 12 2025 at 15:13):

John Baez said:

Now that makes me wonder: which Lawvere theories are equivalent to their opposite category?

I don't know, but I wouldn't be surprised if $\rm Mat(R)$ is in fact the only instance.

John Baez (Jul 12 2025 at 16:09):

That's a fun conjecture! Note the finite products have to be coproducts as well, which is well on the way to being biproducts as they are in $\mathsf{Mat}(R)$ .

Claudio Pisani (Jul 12 2025 at 18:48):

John Baez said:

Note the finite products have to be coproducts as well

... so that an arrow $\,\rm n \to m\,$ is an arrow from a coproduct to a product (of the generating object $x$ ) and so it is a matrix of arrows $x\to x$ ...

Mike Shulman (Jul 12 2025 at 18:57):

Are you assuming that the contravariant autoequivalence is the identity on objects? Or can you prove that?

Claudio Pisani (Jul 12 2025 at 19:51):

Mike Shulman said:

Are you assuming that the contravariant autoequivalence is the identity on objects? Or can you prove that?

well, I was just suggesting an idea of a proof; but maybe the enunciate itself, as you suggest, should be clarified ...

John Baez (Jul 12 2025 at 20:13):

Mike Shulman said:

Are you assuming that the contravariant autoequivalence is the identity on objects? Or can you prove that?

I guess I was stupidly assuming that without noticing. But it might be nice to assume that, see how far one can get with that assumption, and then see what it takes to prove that assumption.

Mike Shulman (Jul 12 2025 at 20:57):

I admit it would be pretty wild if you could have such an autoequivalence that was not the identity on objects.