[contravariant]

I wasn't aware. The examples given here are all cases where the monad doesn't commute, hence my question.

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.

[kmill]

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.

[contravariant]

That's a very interesting article, thanks.

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?

[kmill]

There is nothing at all preventing the existence of multiple distributive laws, and I can think of some (non-programmy) examples that have multiple possibilities. I wouldn't be surprised if some cohomology group classifies them, at least in certain settings.

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).