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by kmill
1579 days ago
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It's not very precise talking about whether a monad commutes -- it's not clear what "commutes" should really mean (certainly not F . G = G . F since that's way too restrictive). I started working it out by hand, but then figured nLab had it somewhere, and indeed: https://golem.ph.utexas.edu/category/2017/02/distributive_la... F, G, and F.G are monads if and only if there is a "distributive law," which is a natural transformation G.F -> F.G satisfying some properties. It's like something that satisfies half the braiding properties, where braidings are already a weaker version of commutativity. |
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That said while FG = GF is indeed restrictive, requiring there to be a natural isomorphism between them is slightly less restrictive and just requiring the existence of a distributive law seems a bit too broad. What's preventing the existence of multiple distributive laws? Is there even anything preventing monads from always having a distributive law?