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Max New (Apr 04 2022 at 09:33):The nlab has a page on monoidal monads and it says
The notion of monoidal monad is equivalent to the notion of commutative monad
However, the two papers I see cited here and here both say that symmetric monoidal monads are equivalent to commutative monads.
Moreover this paper has a monad on a symmetric monoidal category that is monoidal, but not symmetric monoidal, and not commutative, which I found very confusing because I had read the above on the nlab.
So which one is wrong? Is the correct statement only that symmetric monoidal monads are the same as commutative monads?
Morgan Rogers (he/him) (Apr 04 2022 at 10:07):It sounds from your counterexample that the nLab is wrong. It happens! Please do leave a note on the corresponding nForum page so that the page gets edited when that becomes possible again.
Reid Barton (Apr 04 2022 at 10:23):Theorem 12 in that paper doesn't seem to supply an actual example. Not that I checked anything carefully, but my instinct would be to trust the nLab page more (specifically the proof of Proposition 3.1).
Jules Hedges (Apr 04 2022 at 10:25):I'm pretty sure you need a symmetry around to even define what it means for a strong monad to be commutative, but that's not the case for monoidal monads
Max New (Apr 04 2022 at 10:28):Jules Hedges said:
I'm pretty sure you need a symmetry around to even define what it means for a strong monad to be commutative, but that's not the case for monoidal monads
I don't think so, you just seem to need a strength and a costrength
Reid Barton (Apr 04 2022 at 10:31):Right, I think this is the issue--the meaning of "commutative" monad depends on whether it is regarded as a functor between monoidal categories or symmetric monoidal categories
Max New (Apr 04 2022 at 10:31):Reid Barton said:
Theorem 12 in that paper doesn't seem to supply an actual example. Not that I checked anything carefully, but my instinct would be to trust the nLab page more (specifically the proof of Proposition 3.1).
Theorem 12 is based on the syntactic model, so I agree it's not clear that it is not actually commutative. However in section 5 there is a construction for concrete models where this could be checked.
Reid Barton (Apr 04 2022 at 10:31):at least, if you take the union of all the definitions appearing in these sources
Reid Barton (Apr 04 2022 at 10:36):If I have a monad on a symmetric monoidal category but then forget that is symmetric, by the nLab definition, in order to make commutative I get to pick a strength and costrength (though "left" and "right" seem like much better terms) independently. Now since really is symmetric there's an equation that may or may not hold, and it looks plausible that it should be the same as whether or not is symmetric monoidal.
Reid Barton (Apr 04 2022 at 10:43):So everything looks okay to me, though confusing for sure.
Morgan Rogers (he/him) (Apr 04 2022 at 18:00):Thanks for actually checking the source, folks, sorry I was lazy!
Max New (Apr 05 2022 at 21:19):I've reviewed the proofs and now I'm convinced the nlab page is right. If you have a monoidal monad on a monoidal category, the left and right strength commute. And you can show that the monad is necessarily symmetric monoidal as well. So that paper is wrong in saying their monad is not symmetric monoidal. I guess this shows that their models cannot be monadic adjunctions (none of their examples were anyway).
I guess this is a generalization of Eckmann-Hilton?
Naïm Favier (Mar 10 2024 at 09:38):For closure: I've formalised this in Agda here, edited the nLab, and started another thread about the existence of non-symmetric monoidal monads.