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Stream: learning: questions

Topic: Beginner question about cartesian monads

Sándor Bartha (Jan 21 2025 at 15:29):

Why the free commutative monoid monad is not cartesian? It is a (non) example in Tom Leinster: General Operads and Multicategories , Examples 1.4 iii of cartesian monads.

Leinster states that the naturality square for $\mu$ at $2 \to 1$ is not a pullback, but I cannot see why not, it seems to be a pullback to me. But I must have made a mistake in interpreting one of the definitions.

John Baez (Jan 21 2025 at 17:29):

For people are interested, the book is available for free on the arXiv.

Ralph Sarkis (Jan 21 2025 at 17:39):

I put everything in this diagram, but if you don't want to get spoiled, we can start with three simple questions. Let $T$ denote that monad.

Do you see that $T(1)$ is the set of natural numbers $\N$ ?
Do you see that $T(2)$ is the product $\N\times\N$ ?
Do you see that $T(X)$ is the set $\mathcal{M}(X)$ of all [[multisets]] supported in $X$ ?

Sándor Bartha (Jan 21 2025 at 18:06):

Yes, so far these three facts agree with what I have figured out. But I fear I misunderstood one of the arrows on the squere -- I will try to understand your diagram. Thanks!

Ralph Sarkis (Jan 21 2025 at 18:12):

Great! How did you describe the maps in the supposedly pullback square ?
$\mu_2: TT(2) = \mathcal{M}(\N \times \N) \to \N \times \N = T(2)$
$\mu_1: TT(1) = \mathcal{M}(\N) \to \N = T(1)$
$T!: \N \times \N \to \N$
$TT!: \mathcal{M}(\N\times \N) \to \mathcal{M}(\N)$

Sándor Bartha (Jan 21 2025 at 18:46):

Well, all of them sums something:
$\mu_2(m) = (\sum_{(x,y) \in m} x , \sum_{(x,y)\in m} y)$
$\mu_1(m) = \sum_{x \in m} x$
$T!(x,y) = x +y$
$TT!(m)=\{ x+y \mid (x,y) \in m \}$

I am not sure about multiset notation, but all of the variables in the sums and the set interpreted as multiset of course.

Sándor Bartha (Jan 21 2025 at 18:56):

Ah, I see your counterexample -- now that I see it, it is kind of hard to remember what I missed earlier. Thank you for the wonderful diagram and the extremly quick help!

I tried to get some intuition why the free monoid monad is cartesian while the commutative is not, and I got confused at some point. Now I continue my journey and try to understand why it is important for a monad to be cartesian.