5 ms·

"Monads don't compose" is a common sentence (google it!), for example here: https://stackoverflow.com/questions/7040844/applicatives-compose-monads-dont https:/

"Monads don't compose" is a common sentence (google it!), for example here: https://stackoverflow.com/questions/7040844/applicatives-compose-monads-dont https://stackoverflow.com/questions/7040844/applicatives-com... Maybe you already knew about this and is still asking, though?

contravariant 5y ago
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 5y ago
  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_laws.html 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 5y ago
    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 5y ago
      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).

contravariant 5y ago

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 5y ago
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_laws.html 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 5y ago
  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 5y ago
    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).