"Monads don't compose" is a common sentence (google it!), for example here: https://stackoverflow.com/questions/7040844/applicatives-com...
Maybe you already knew about this and is still asking, though?
Maybe you already knew about this and is still asking, though?
If two monads commute you can show that the composition of the two is (trivially) a new monad, but I'm not sure if the converse also holds.
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.
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?
I don't know about whether distributive laws always exist, but what is probably true is that there's no universal distributive law -- that's in the sense that you have a function from pairs of monads to distributive laws that is natural with respect to homomorphisms of monads, whatever those are (I know there's a bicategory of monads, but that's about it).