Algebraically interpreting polymorphism

Question

So I understand the basic algebraic interpretation of types:

Either a b ~ a + b
(a, b) ~ a * b
a -> b ~ b^a
()   ~ 1
Void ~ 0 -- from Data.Void

... and that these relations are true for concrete types, like Bool, as opposed to polymorphic types like a. I also know how to translate type signatures with polymorphic types into their concrete type representations by just translating the Church encoding according to the following isomorphism:

(forall r . (a -> r) -> r) ~ a

So if I have:

id :: forall a . a -> a

I know that it does not mean id ~ a^a, but it actually means:

id :: forall a . (() -> a) -> a
id ~ ()
   ~ 1

Similarly:

pair :: forall r . (a -> b -> r) -> r
pair ~ ((a, b) -> r) - > r
     ~ (a, b)
     ~ a * b

Which brings me to my question. What is the "algebraic" interpretation of this rule:

(forall r . (a -> r) -> r) ~ a

For every concrete type isomorphism I can point to an equivalent algebraic rule, such as:

(a, (b, c)) ~ ((a, b), c)
a * (b * c) = (a * b) * c

a -> (b -> c) ~ (a, b) -> c
(c^b)^a = c^(b * a)

But I don't understand the algebraic equality that is analogous to:

(forall r . (a -> r) -> r) ~ a

Answer 1

This is the famous Yoneda lemma for the identity functor.

Check this post for a readable introduction, and any category theory textbook for more.

Briefly, given f :: forall r. (a -> r) -> r you can apply f id to get an a, and conversely, given x :: a you can take ($x) to get forall r. (a -> r) -> r.

These operations are mutually inverse. Proof:

Obviously ($x) id == x. I will show that

($(f id)) == f,

since functions are equal when they are equal on all arguments, let's take x :: a -> r and show that

($(f id)) x == f x i.e.

x (f id) == f x.

Since f is polymorphic, it works as a natural transformation; this is the naturality diagram for f:

               f_A
     Hom(A, A)  →  A
(x.)      ↓        ↓ x
     Hom(A, R)  →  R
               f_R

So x . f == f . (x.).

Plugging identity, (x . f) id == f x. QED

Answer 2

(Rewritten for clarity)

There seem to be two parts to your question. One is implied and is asking what the algebraic interpretation of forall is, and the other is asking about the cont/Yoneda transformation, which sdcvvc's answer already covered pretty well.

I'll try to address the algebraic interpretation of forall for you. You mention that A -> B is B^A but I'd like to take that a step further and expand it out to B * B * B * ... * B (|A| times). Although we do have exponentiation as a notation for repeated multiplication like that, there's a more flexible notation, ∏ (uppercase Pi) representing arbitrary indexed products. There are two components to a Pi: the range of values we want to multiply over, and the expression that we're multiplying out. For example, at the value level, you might express the factorial function as fact i = ∏ [1..i] (λx -> x).

Going back to the world of types, we can view the exponentiation operator in the A -> B ~ B^A correspondence as a Pi: B^A ~ ∏ A (λ_ -> B). This says that we're defining an A-ary product of Bs, such that the Bs cannot depend on the particular A we've chosen. Sure, it's equivalent to plain exponentiation, but it lets us move up to cases in which there is a dependence.

In the most general case, we get dependent types, like what you see in Agda or Coq: in Agda syntax, replicate : Bool -> ((n : Nat) -> Vec Bool n) is one possible application of a Pi type, which could be expressed more explicitly as replicate : Bool -> ∏ Nat (Vec Bool), or further as replicate : ∏ Bool (λ_ -> ∏ Nat (Vec Bool)).

Note that as you might expect from the underlying algebra, you can fuse both of the ∏s in the definition of replicate above into a single ∏ ranging over the cartesian product of the domains: ∏ Bool (\_ -> ∏ Nat (Vec Bool)) is equivalent to ∏ (Bool, Nat) (λ(_, n) -> Vec Bool n) just like it would be at the "value level". This is simply uncurrying from the perspective of type theory.

I do realize your question was about polymorphism, so I'll stop going on about dependent types, but they are relevant: forall in Haskell is roughly equivalent to a ∏ with a domain over the type (kind) of types, *. Indeed, the function-like behavior of polymorphism can be observed directly in GHC core, which types them as capital lambdas (Λ). As such, a polymorphic type like forall a. a -> a is actually just ∏ * (Λ a -> (a -> a)) (using the Λ notation now that we distinguish between types and values), which can be expanded out to the infinite product (Bool -> Bool, Int -> Int, () -> (), (Int -> Bool) -> (Int -> Bool), ...) for every possible type. Instantiation of the type variable is simply projecting out the suitable element from the *-ary product (or applying the type function).

Now, for the big piece I missed in my original version of this answer: parametricity. Parametricity can be described in several different ways, but none of the ones I know of (viewing types as relations, or (di)naturality in category theory) really has a very algebraic interpretation. For our purposes, though, it boils down to something fairly simple: you can't pattern-match on *. I know that GHC lets you do that at the type level with type families, but you can only cover a finite chunk of * when doing that, so there are necessarily always points at which your type family is undefined.

What this means, from the point of view of polymorphism, is that any type function F we write in ∏ * F must either be constant (i.e., completely ignore the type it was polymorphic over) or pass the type through unchanged. Thus, ∏ * (Λ _ -> B) is valid because it ignores its argument, and corresponds to forall a. B. The other case is something like ∏ * (Λ x -> Maybe x), which corresponds to forall a. Maybe a, which doesn't ignore the type argument, but only "passes it through". As such, a ∏ A that has an irrelevant domain A (such as when A = *) can be seen as more of an A-ary indexed intersection (picking the common elements across all instantiations of the index), rather than a product.

Crucially, at the value level, the rules of parametricity prevent any funny behavior that might suggest the types are larger than they really are. Because we don't have typecase, we can't construct a value of type forall a. B that does something different based on what a was instantiated to. Thus, although the type is technically a function * -> B, it is always a constant function, and is thus equivalent to a single value of B. Using the ∏ interpretation, it is indeed equivalent to an infinite *-ary product of Bs, but those B values must always be identical, so the infinite product is effectively as big as a single B.

Similarly, although ∏ * (Λ x -> (x -> x)) (a.k.a., forall a. a -> a) is technically equivalent to an infinite product of functions, none of those functions can inspect the type, so all are constrained to only return their input value and not do any funny business like (+1) : Int -> Int when instantiated to Int. Because there is only one (assuming a total language) function that can't inspect the type of its argument but must return a value of that same type, the infinite product is thus just as large as a single value.

Now, about your direct question on (forall r . (a -> r) -> r) ~ a. First, let's express your ~ operator more formally. It's really isomorphism, so we need two functions going back and forth, and an argument that they're inverses.

data Iso a b = Iso 
  { to   :: a -> b
  , from :: b -> a
  -- proof1 :: forall x. to (from x) == x
  -- proof2 :: forall x. from (to x) == x
  }

and now we express your original question in more formal terms. Your question amounts to constructing a term of the following (impredicative, so GHC has trouble with it, but we'll survive) type:

forall a. Iso (forall r. (a -> r) -> r) a

Which, using my earlier terminology, amounts to ∏ * (Λ a -> Iso (∏ * (Λ r -> ((a -> r) -> r))) a). Once again we have an infinite product that can't inspect its type argument. By handwaving, we can argue that the only possible values considering the parametricity rules (the other two proofs are respected automatically) for to and from are ($ id) and flip id.

If this feels unsatisfying, it's probably because the algebraic interpretation of forall didn't really add anything to the proof. It's really just plain old type theory, but I hope I was able to provide something that feels a little less categorical than the Yoneda form of it. It's worth noting that we don't actually need to use parametricity to write proof1 and proof2 above, though. Parametricity only enters the picture when we want to state that ($ id) and flip id are our only options for to and from (which we can't prove in Agda or Coq, for that reason).

Answer 3

To (attempt to) answer the actual question (which is less interesting than the answers to the broader issues raised), the question is ill formed because of a "type error"

Either ~ (+) 
(,)    ~ (*)
(->) b ~ flip (^)
()   ~ 1
Void ~ 0 

These all map types to integers, and type constructors to functions on naturals. In a sense, you have a functor from the category of types to the category of naturals. In the other direction, you "forget" stuff, since the types preserve algebraic structure while the naturals throw it away. I.e. given Either () () you can get a unique natural, but given that natural, you can get many types.

But this is different:

(forall r . (a -> r) -> r) ~ a

It maps a type to another type! It is not part of the above functor. It's just an isomorphism within the category of types. So let's give that a different symbol, <=>

Now we have

(forall r . (a -> r) -> r) <=> a

Now you note that we can not only send types to nats and arrows to arrows, but also some isomorphisms to other isomorphisms:

(a, (b, c)) <=> ((a, b), c) ~ a * (b * c) = (a * b) * c

But something subtle is going on here. In a sense, the latter isomorphism on pairs is true because the algebraic identity is true. This is to say that the "isomorphism" in the latter simply means that the two types are equivalent under the image of our functor to the nats.

The former isomorphism we need to prove directly, which is where we start to get to the underlying question -- is given our functor to the nats, what does forall r. map to? But the answer is that forall r. is neither a type, nor a meaningful arrow between types.

By introducing forall, we have moved away from first order types. There's no reason to expect that forall should fit in our above Functor, and indeed, it doesn't.

So we can explore, as others have above, why the isomorphism holds (which is itself very interesting) -- but in doing so we've abandoned the algebraic core of the question. A question which can be answered, I think, is, given the category of higher-order types and constructors as arrows between them, what is there meaningful Functor to?

Edit: So now I have another approach which shows why adding polymorphism makes things go nuts. We start by asking a simpler question -- does a given polymorphic type have zero or more than zero inhabitants? This is the type inhabitation problem, and winds up being, via Curry-Howard, a problem in modified realizability, since it's the same thing as asking if a formula in some logic is realizable in an appropriate computational model. Now as that page explains, this is decidable in the simply typed lambda calculus but is PSPACE-complete. But once we move to anything more complicated, by adding polymorphism for example and going to System F, then it goes to undecidable!

So, if we can't decide if an arbitrary type is inhabited at all, then we clearly can't decide how many inhabitants it has!

Answer 4

It's an interesting question. I don't have a full answer, but this was too long for a comment.

The type signature (forall r. (a -> r) -> r) can be expressed as me saying

For any type r that you care to name, if you give me a function that takes a and produces an r, then I will give you back an r.

Now, this has to work for any type r, but it can be a specific type a. So the way for me to pull of this neat trick is to have an a sitting around somewhere, that I feed to the function (which produces an r for me) and then I hand that r back to you.

But if I have an a sitting around, I could give it to you:

If you give me a 1, I'll give you an a.

which corresponds to the type signature 1 -> a or simply a. By this informal argument we have

(forall r. (a -> r) -> r) ~ a

The next step would be to generate the corresponding algebraic expression, but I'm not clear on how the algebraic quantities interact with the universal quantification. We may need to wait for an expert!

Answer 5

A few links to the nLab:

• Universal quantifier, corresponds to dependent product.

• Existential quantifier, corresponds to dependent sum (dependent coproduct).

---

Thus, in settings of category theory:

Type               | Modeled¹ as               | In category
-------------------+---------------------------+-------------
Unit               | Terminal object           | CCC
Bottom             | Initial object            |
Record             | Product                   |
Union              | Sum (coproduct)           |
Function           | Exponential               |
-------------------+---------------------------+-------------
Dependent product² | Right adjoint to pullback | LCCC
Dependent sum      | Left adjoint to pullback  |

¹) in appropriate category ─ CCC for total and non-polymorphic subset of Haskell (link), CPO for non-total traits of Haskell (link), LCCC for dependently typed languages.

²) forall quantification is a special case of dependent product:

∀(x :: *). y[x] ~ ∏(x : Set)y[x]

where Set is the universe of all small types.