Adjoint functors determine monad transformers, but where's lift?

问题

I'm intrigued by the construction described here for determining a monad transformer from adjoint functors. Here's some code that summarizes the basic idea:

{-# LANGUAGE MultiParamTypeClasses #-}

import           Control.Monad

newtype Three g f m a = Three { getThree :: g (m (f a)) }

class (Functor f, Functor g) => Adjoint f g where
  counit :: f (g a) -> a
  unit   :: a -> g (f a)

instance (Adjoint f g, Monad m) => Monad (Three g f m) where
  return  = Three . fmap return . unit
  m >>= f = Three $ fmap (>>= counit . fmap (getThree . f)) (getThree m)

instance (Adjoint f g, Monad m) => Applicative (Three g f m) where
  pure  = return
  (<*>) = ap

instance (Adjoint f g, Monad m) => Functor (Three g f m) where
  fmap = (<*>) . pure

Given that Adjoint ((,) s) ((->) s), Three ((->) s) ((,) s) appears equivalent to StateT s.

Very cool, but I am puzzled by a couple things:

• How can we upgrade a monadic m a into a monadic Three g f m a? For the specific case of Three ((->) s) ((,) s), it's of course obvious how to do this, but it seems desirable to have a recipe that works for any Three g f provided that Adjoint f g. In other words, it seems like there should be an analog of lift whose definition only requires unit, counit, and the return and >>= of the input monad. But I cannot seem to find one (I have seen a definition using sequence, but this seems a bit like cheating since it requires f to be Traversable).

• For that matter, how can we upgrade g a into a Three g f m a (provided Adjoint f g)? Again, for the specific case of Three ((->) s) ((,) s) it's obvious how to do this, but I'm wondering if there's an analog of gets that only requires unit, counit, and the return and >>= of the input monad.

回答1:

lift, in Benjamin Hodgson's answer, is set up as:

lift mx = let mgfx = fmap unit mx
              gmfx = distributeR mgfx
          in Three gmfx
-- or
lift = Three . distributeR . fmap unit

As you know, that is not the only plausible strategy we might use there:

lift mx = let gfmx = unit mx
              gmfx = fmap sequenceL gfmx
          in Three gmfx
-- or
lift = Three . fmap sequenceL . unit

Whence the Traversable requirement for Edward Kmett's corresponding MonadTrans instance originates. The question, then, becomes whether relying on that is, as you put it, "cheating". I am going to argue it is not.

We can adapt Benjamin's game plan concerning Distributive and right adjoints and try to find whether left adjoints are Traversable. A look at Data.Functor.Adjunction shows we have a quite good toolbox to work with:

unabsurdL :: Adjunction f u => f Void -> Void
cozipL :: Adjunction f u => f (Either a b) -> Either (f a) (f b)
splitL :: Adjunction f u => f a -> (a, f ())
unsplitL :: Functor f => a -> f () -> f a

Edward helpfully tells us that unabsurdL and cozipL witness that "[a] left adjoint must be inhabited, [and that] a left adjoint must be inhabited by exactly one element", respectively. That, however, means splitL corresponds precisely to the shape-and-contents decomposition that characterises Traversable functors. If we add to that the fact that splitL and unsplitL are inverses, an implementation of sequence follows immediately:

sequenceL :: (Adjunction f u, Functor m) => f (m a) -> m (f a)
sequenceL = (\(mx, fu) -> fmap (\x -> unsplitL x fu) mx) . splitL

(Note that no more than Functor is demanded of m, as expected for traversable containers that hold exactly one value.)

All that is missing at this point is verifying that both implementations of lift are equivalent. That is not difficult, only a bit laborious. In a nutshell, the distributeR and sequenceR definitions here can be simplified to:

distributeR = \mgx ->
    leftAdjunct (\fa -> fmap (\gx -> rightAdjunct (const gx) fa) mgx) ()   
sequenceL =
    rightAdjunct (\mx -> leftAdjunct (\fu -> fmap (\x -> fmap (const x) fu) mx) ())

We want to show that distributeR . fmap unit = fmap sequenceL . unit. After a few more rounds of simplification, we get:

distributeR . fmap unit = \mgx ->
    leftAdjunct (\fa -> fmap (\gx -> rightAdjunct (const (unit gx)) fa) mgx) ()
fmap sequenceL . unit = \mx ->
    leftAdjunct (\fu -> fmap (\x -> fmap (const x) fu) mx) ()

We can show those are really the same thing by picking \fu -> fmap (\x -> fmap (const x) fu) mx -- the argument to leftAdjunct in the second right-hand side -- and slipping rightAdjunct unit = counit . fmap unit = id into it:

\fu -> fmap (\x -> fmap (const x) fu) mx
\fu -> fmap (\x -> fmap (const x) fu) mx
\fu -> fmap (\x -> (counit . fmap unit . fmap (const x)) fu) mx
\fu -> fmap (\x -> rightAdjunct (unit . const x) fu) mx
\fu -> fmap (\x -> rightAdjunct (const (unit x)) fu) mx
-- Sans variable renaming, the same as
-- \fa -> fmap (\gx -> rightAdjunct (const (unit gx)) fa) mgx

The takeaway is that the Traversable route towards your MonadTrans is just as secure as the Distributive one, and concerns about it -- including the ones mentioned by the Control.Monad.Trans.Adjoint documentation -- should no longer trouble anyone.

P.S.: It is worth noting that the definition of lift put forward here can be spelled as:

lift = Three . leftAdjunct sequenceL

That is, lift is sequenceL sent through the adjunction isomorphism. Additionally, from...

leftAdjunct sequenceL = distributeR . fmap unit

... if we apply rightAdjunct on both sides, we get...

sequenceL = rightAdjunct (distributeR . fmap unit)

... and if we compose fmap (fmap counit) on the left of both sides, we eventually end up with:

distributeR = leftAdjunct (fmap counit . sequenceL)

So distributeR and sequenceL are interdefinable.

回答2:

How can we upgrade a monadic m a into a monadic Three g f m a?

Good question. Time for a game of type tennis!

-- i'm using Adjuction from the adjunctions package because I'll need the fundeps soon
lift :: Adjunction f g => m a -> Three g f m a
lift mx = Three _

The hole is typed g (m (f a)). We have mx :: m a in scope, and of course unit :: a -> g (f a) and fmap :: (a -> b) -> m a -> m b.

lift mx = let mgfx = fmap unit mx
          in Three $ _ mgfx

Now it's _ :: m (g (f a)) -> g (m (f a)). This is distribute if g is Distributive.

lift mx = let mgfx = fmap unit mx
              gmfx = distributeR mgfx
          in Three gmfx
-- or
lift = Three . distributeR . fmap unit

So now we just need to prove that the right hand side of an adjunction is always Distributive:

distributeR :: (Functor m, Adjunction f g) => m (g x) -> g (m x)
distributeR mgx = _

Since we need to return a g, the clear choice of methods from Adjunction is leftAdjunct :: Adjunction f g => (f a -> b) -> a -> g b, which uses unit to create a g (f a) and then tears down the inner f a by fmapping a function.

distributeR mgx = leftAdjunct (\fa -> _) _

I'm going to attack the first hole first, with the expectation that filling it in might tell me something about the second one. The first hole has a type of m a. The only way we can get hold of an m of any type is by fmapping something over mgx.

distributeR mgx = leftAdjunct (\fa -> fmap (\gx -> _) mgx) _

Now the first hole has a type of a, and we have gx :: g a in scope. If we had an f (g a) we could use counit. But we do have an f x (where x is currently an ambiguous type variable) and a g a in scope.

distributeR mgx = leftAdjunct (\fa -> fmap (\gx -> counit (fa $> gx)) mgx) _

It turns out that the remaining hole has an ambiguous type, so we can use anything we want. (It'll be ignored by $>.)

distributeR mgx = leftAdjunct (\fa -> fmap (\gx -> counit (fa $> gx)) mgx) ()

---

That derivation may have looked like a magic trick but really you just get better at type tennis with practice. The skill of the game is being able to look at the types and apply intuitions and facts about the objects you're working with. From looking at the types I could tell that I was going to need to exchange m and g, and traversing m was not an option (because m is not necessarily Traversable), so something like distribute was going to be necessary.

Besides guessing I was going to need to implement distribute, I was guided by some general knowledge about how adjunctions work.

Specifically, when you're talking about * -> *, the only interesting adjunctions are (uniquely isomorphic to) the Reader/Writer adjunction. In particular, that means any right adjoint on Hask is always Representable, as witnessed by tabulateAdjunction and indexAdjunction. I also know that all Representable functors are Distributive (in fact logically the converse is also true, as described in Distributive's docs, even though the classes aren't equivalent in power), per distributeRep.

---

For that matter, how can we upgrade g a into a Three g f m a (provided Adjoint f g)?

I'll leave that as an exercise. I suspect you'll need to lean on the g ~ ((->) s) isomorphism again. I actually don't expect this one to be true of all adjunctions, just the ones on Hask, of which there is only one.

来源：https://stackoverflow.com/questions/49322276/adjoint-functors-determine-monad-transformers-but-wheres-lift