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I am working on a theory of particle physics where we use monads. I have a few conjectures that I need to check.

  1. The category of $\mathbb{C}$-modules is monadic over set
  2. The category of $\mathbb{N}$-modules is monadic over Set
  3. Since the monad, $\mathcal{M}_C$, which factors through the category of $\mathbb{C}$-modules is seated on Set and likewise for the monad, $\mathcal{M}_N$, that factors through the category of $\mathbb{N}$-modules, these monads compose according to functor composition. Their composition, $\mathcal{M}_P = \mathcal{M}_C \cdot \mathcal{M}_N$, is also a monad on SET.

The physical interpretation: The monad, $\mathcal{M}_P$, can be interpreted as complex combinations of collections of particle types. Thus, this monad can be used to represent quantum states for the input/output of a particle scattering experiment.

Edit:

Someone has pointed out a very important fact, which is that I have to specify a distributive law for the functor composition to become a monad. Unfortunately, I don't know how to define this. Perhaps the question can be phrased better as follows. Can we define a distributive law to make the composition into a monad? What is the distributive law? If there is a large freedom, perhaps we can select a law that captures some aspect of the physics we are trying to model.

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    $\begingroup$ You can’t compose monads and get another monad- you have to specify a ‘distributive’ law between them to endow the composite with the structure of a monad. $\endgroup$ Commented May 30, 2019 at 21:53
  • $\begingroup$ What is an N-module here? $\endgroup$ Commented Oct 29, 2023 at 20:50
  • $\begingroup$ @MartinBrandenburg Group carrying an N-action appropriately compatible with the group structure would obviously be the naive guess; does this not work out for some reason? $\endgroup$
    – Alec Rhea
    Commented Oct 30, 2023 at 4:37
  • $\begingroup$ @MartinBrandenburg I think I mean a semimodule over the natural numbers. That's what I mean by N-module. $\endgroup$
    – Ben Sprott
    Commented Nov 8, 2023 at 15:00
  • $\begingroup$ Every commutative monoid is uniquely an $\mathbb{N}$-semimodule in the same way that an abelian group is a $\mathbb {Z}$-module. $\endgroup$
    – Ben Sprott
    Commented Nov 8, 2023 at 15:08

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