module Algebra.Group.Ab.Tensor where

Bilinear maps🔗

A function where all types involved are equipped with abelian group structures, is called bilinear when it satisfies and it is a group homomorphism in each of its arguments.

record Bilinear (A : Abelian-group ) (B : Abelian-group ℓ') (C : Abelian-group ℓ'') : Type (  ℓ'  ℓ'') where
  private
    module A = Abelian-group-on (A .snd)
    module B = Abelian-group-on (B .snd)
    module C = Abelian-group-on (C .snd)

  field
    map     :  A    B    C 
    pres-*l :  x y z  map (x A.* y) z  map x z C.* map y z
    pres-*r :  x y z  map x (y B.* z)  map x y C.* map x z
  fixl-is-group-hom :  a 
    is-group-hom B.Abelian→Group-on C.Abelian→Group-on (map a)
  fixl-is-group-hom a .is-group-hom.pres-⋆ x y = pres-*r a x y

  fixr-is-group-hom :  b 
    is-group-hom A.Abelian→Group-on C.Abelian→Group-on  a  map a b)
  fixr-is-group-hom b .is-group-hom.pres-⋆ x y = pres-*l x y b

  module fixl {a} = is-group-hom (fixl-is-group-hom a)
  module fixr {a} = is-group-hom (fixr-is-group-hom a)

  open fixl
    renaming ( pres-id   to pres-idr
             ; pres-inv  to pres-invr
             ; pres-diff to pres-diffr
             )
    hiding ( pres-⋆ )
    public
  open fixr
    renaming ( pres-id   to pres-idl
             ; pres-inv  to pres-invl
             ; pres-diff to pres-diffl
             )
    hiding ( pres-⋆ )
    public

module _ {A : Abelian-group } {B : Abelian-group ℓ'} {C : Abelian-group ℓ''} where
  private
    module A = Abelian-group-on (A .snd)
    module B = Abelian-group-on (B .snd)
    module C = Abelian-group-on (C .snd)

    Bilinear-path
      : {ba bb : Bilinear A B C}
       (∀ x y  Bilinear.map ba x y  Bilinear.map bb x y)
       ba  bb
    Bilinear-path {ba = ba} {bb} p = q where
      open Bilinear

      q : ba  bb
      q i .map x y = p x y i
      q i .pres-*l x y z = is-prop→pathp  i  C.has-is-set (p (x A.* y) z i) (p x z i C.* p y z i))
        (ba .pres-*l x y z) (bb .pres-*l x y z) i
      q i .pres-*r x y z = is-prop→pathp  i  C.has-is-set (p x (y B.* z) i) (p x y i C.* p x z i))
        (ba .pres-*r x y z) (bb .pres-*r x y z) i

  instance
    Extensional-bilinear
      :  {ℓr}  ef : Extensional ( A    B    C ) ℓr   Extensional (Bilinear A B C) ℓr
    Extensional-bilinear  ef  = injection→extensional!  h  Bilinear-path  x y  h · x · y)) ef

module _ {} {A B C : Abelian-group } where

We have already noted that the set of group homomorphisms between a pair of abelian groups is an abelian group, under pointwise multiplication. The type of bilinear maps is equivalent to the type of group homomorphisms

  curry-bilinear : Bilinear A B C  Ab.Hom A Ab[ B , C ]
  curry-bilinear f .fst a .fst = f .Bilinear.map a
  curry-bilinear f .fst a .snd = Bilinear.fixl-is-group-hom f a
  curry-bilinear f .snd .is-group-hom.pres-⋆ x y = ext λ z 
    f .Bilinear.pres-*l _ _ _

  curry-bilinear-is-equiv : is-equiv curry-bilinear
  curry-bilinear-is-equiv = is-iso→is-equiv morp where
    morp : is-iso curry-bilinear
    morp .is-iso.from uc .Bilinear.map x y = uc · x · y
    morp .is-iso.from uc .Bilinear.pres-*l x y z = ap ( _) (uc .snd .is-group-hom.pres-⋆ _ _)
    morp .is-iso.from uc .Bilinear.pres-*r x y z = (uc · _) .snd .is-group-hom.pres-⋆ _ _
    morp .is-iso.rinv uc = ext λ _ _  refl
    morp .is-iso.linv uc = ext λ _ _  refl

The tensor product🔗

Thinking about the currying isomorphism we set out to search for an abelian group which lets us “associate” Bilinear in the other direction: Bilinear maps are equivalent to group homomorphisms but is there a construction “”, playing the role of product type, such that is also the type of bilinear maps

The answer is yes: rather than we write this infix as and refer to it as the tensor product of abelian groups. Since is determined by the maps out of it, we can construct it directly as a higher inductive type. We add a constructor for every operation we want, and for the equations these should satisfy: should be an Abelian group, and it should admit a bilinear map

  data Tensor : Type (  ℓ') where
    :1       : Tensor
    _:*_     : Tensor  Tensor  Tensor
    :inv     : Tensor  Tensor
    squash   : is-set Tensor
    t-invl   :  {x}  :inv x :* x  :1
    t-idl    :  {x}  :1 :* x  x
    t-assoc  :  {x y z}  x :* (y :* z)  (x :* y) :* z
    t-comm   :  {x y}  x :* y  y :* x

    _,_       :  A    B   Tensor
    t-pres-*r :  {x y z}  (x , y B.* z)  (x , y) :* (x , z)
    t-pres-*l :  {x y z}  (x A.* y , z)  (x , z) :* (y , z)

The first 8 constructors are, by definition, exactly what we need to make Tensor into an abelian group. The latter three are the bilinear map Since this is an inductive type, it’s the initial object equipped with these data.

  open make-abelian-group
  make-abelian-tensor : make-abelian-group Tensor
  make-abelian-tensor .ab-is-set   = squash
  make-abelian-tensor .mul         = _:*_
  make-abelian-tensor .inv         = :inv
  make-abelian-tensor .1g          = :1
  make-abelian-tensor .idl x       = t-idl
  make-abelian-tensor .assoc x y z = t-assoc
  make-abelian-tensor .invl x      = t-invl
  make-abelian-tensor .comm x y    = t-comm

  _⊗_ : Abelian-group (  ℓ')
  _⊗_ = to-ab make-abelian-tensor

  to-tensor : Bilinear A B _⊗_
  to-tensor .Bilinear.map           = _,_
  to-tensor .Bilinear.pres-*l x y z = t-pres-*l
  to-tensor .Bilinear.pres-*r x y z = t-pres-*r

If we have any bilinear map we can first extend it to a function of sets by recursion, then prove that this is a group homomorphism. It turns out that this extension is definitionally a group homomorphism.

  bilinear-map→function : Bilinear A B C  Tensor A B   C 
  bilinear-map→function bilin = go module bilinear-map→function where
    go : Tensor A B   C 
    go :1       = C.1g
    go (x :* y) = go x C.* go y
    go (:inv x) = go x C.⁻¹
    go (x , y)  = Bilinear.map bilin x y

    go (squash x y p q i j)      = C.has-is-set (go x) (go y)  i  go (p i))  i  go (q i)) i j
    go (t-invl {x} i)            = C.inversel {x = go x} i
    go (t-idl {x} i)             = C.idl {x = go x} i
    go (t-assoc {x} {y} {z} i)   = C.associative {x = go x} {go y} {go z} i
    go (t-comm {x} {y} i)        = C.commutes {x = go x} {y = go y} i
    go (t-pres-*r {a} {b} {c} i) = Bilinear.pres-*r bilin a b c i
    go (t-pres-*l {a} {b} {c} i) = Bilinear.pres-*l bilin a b c i

  {-# DISPLAY bilinear-map→function.go b = bilinear-map→function b #-}
  from-bilinear-map : Bilinear A B C  Ab.Hom (A  B) C
  from-bilinear-map bilin .fst = bilinear-map→function A B C bilin
  from-bilinear-map bilin .snd .is-group-hom.pres-⋆ x y = refl

It’s also easy to construct a function in the opposite direction, turning group homomorphisms into bilinear maps, but proving that this is an equivalence requires appealing to an induction principle of Tensor which handles the equational fields: Tensor-elim-prop.

  to-bilinear-map : Ab.Hom (A  B) C  Bilinear A B C
  to-bilinear-map gh .Bilinear.map x y = gh · (x , y)
  to-bilinear-map gh .Bilinear.pres-*l x y z =
    ap (apply gh) t-pres-*l  gh .snd .is-group-hom.pres-⋆ _ _
  to-bilinear-map gh .Bilinear.pres-*r x y z =
    ap (apply gh) t-pres-*r  gh .snd .is-group-hom.pres-⋆ _ _

  from-bilinear-map-is-equiv : is-equiv from-bilinear-map
  from-bilinear-map-is-equiv = is-iso→is-equiv morp where
    morp : is-iso from-bilinear-map
    morp .is-iso.from = to-bilinear-map
    morp .is-iso.rinv hom = ext $ Tensor-elim-prop A B  x  C.has-is-set _ _)
       x y  refl)
       x y  ap₂ C._*_ x y  sym (hom .snd .is-group-hom.pres-⋆ _ _))
       x  ap C._⁻¹ x  sym (is-group-hom.pres-inv (hom .snd)))
      (sym (is-group-hom.pres-id (hom .snd)))
    morp .is-iso.linv x = ext λ _ _  refl

The tensor-hom adjunction🔗

Since we have a construction satisfying we’re driven, being category theorists, to question its naturality: Is the tensor product a functor, and is this equivalence of Hom-sets an adjunction?

The answer is yes, and the proofs are essentially plugging together existing definitions, other than the construction of the functorial action of

Ab-tensor-functor : Functor (Ab  ×ᶜ Ab ) (Ab )
Ab-tensor-functor .F₀ (A , B) = A  B
Ab-tensor-functor .F₁ (f , g) = from-bilinear-map bilin where
  bilin : Bilinear _ _ _
  bilin .Bilinear.map x y       = f · x , g · y
  bilin .Bilinear.pres-*l x y z = ap (_, g · z) (f .snd .is-group-hom.pres-⋆ _ _)  t-pres-*l
  bilin .Bilinear.pres-*r x y z = ap (f · x ,_) (g .snd .is-group-hom.pres-⋆ _ _)  t-pres-*r
Ab-tensor-functor .F-id    = ext λ _ _  refl
Ab-tensor-functor .F-∘ f g = ext λ _ _  refl

Tensor⊣Hom : (A : Abelian-group )  Bifunctor.Left Ab-tensor-functor A  Bifunctor.Right Ab-hom-functor A
Tensor⊣Hom A = hom-iso→adjoints to to-eqv nat where
  to :  {x y}  Ab.Hom (x  A) y  Ab.Hom x Ab[ A , y ]
  to f = curry-bilinear (to-bilinear-map f)

  to-eqv :  {x y}  is-equiv (to {x} {y})
  to-eqv = ∘-is-equiv curry-bilinear-is-equiv (Hom≃Bilinear .snd)

  nat : hom-iso-natural {L = Bifunctor.Left Ab-tensor-functor A} {R = Bifunctor.Right Ab-hom-functor A} to
  nat f g h = ext λ _ _  refl

As a monoidal category🔗

We can construct associators and unitors for the tensor product and show that these are coherent, thus making into a monoidal category. While the construction is tedious, it is not complicated. We start with the associator, which, componentwise, is given by sending the triple to We have to show that this is linear in every variable to construct this map, but since we’re simply mapping back into a tensor product, this is by construction.

private
  assc : Associator-for {O = }  _ _  Ab ) Ab-tensor-functor
  assc = to-natural-iso mk where
    mk : make-natural-iso _ _
    mk .eta (G , H , I) = R-adjunct (Tensor⊣Hom _) $ from-bilinear-map λ where
      .map g h  ∫hom  i  g , (h , i))
        record { pres-⋆ = λ x y  ap₂ Tensor._,_ refl t-pres-*r  t-pres-*r }
      .pres-*l x y z  ext λ i  t-pres-*l  refl
      .pres-*r x y z  ext λ i  ap₂ Tensor._,_ refl t-pres-*l  t-pres-*r

    mk .inv (G , H , I) = R-adjunct (Tensor⊣Hom _) record where
      fst g = from-bilinear-map λ where
        .map h i  (g , h) , i
        .pres-*l x y z  ap₂ Tensor._,_ t-pres-*r refl  t-pres-*l
        .pres-*r x y z  t-pres-*r
      snd = record where
        pres-⋆ x y = ext λ h i  ap₂ Tensor._,_ t-pres-*l refl  t-pres-*l

In what will become a theme, the proofs that these constructions are natural inverses are all by trivial computations.

    mk .eta∘inv _     = ext λ _ _ _  refl
    mk .inv∘eta _     = ext λ _ _ _  refl
    mk .natural x y f = ext λ _ _ _  refl

Let us now construct the unit, and the unitors. Recall that the group of integers is the free (Abelian) group on one generator, i.e. that we have a natural equivalence between elements of and maps We will take as the tensor unit. The left unitor sends to the pair To construct a map in the other direction, observe that it suffices to give a for which it suffices to give any the only natural choice is the identity map.

Ab-monoidal : Monoidal-category (Ab )
Ab-monoidal .-⊗-  = Ab-tensor-functor
Ab-monoidal .Unit = Lift-ab _ ℤ-ab

Ab-monoidal .unitor-l = to-natural-iso λ where
  .eta G  ∫hom  x  1 , x) record { pres-⋆ = λ x y  t-pres-*r }

  .inv G  R-adjunct (Tensor⊣Hom G)
    let
      h : Groups.Hom (Lift-group _ ) (Abelian→Group Ab[ G , G ])
      h = pow-hom (Abelian→Group Ab[ G , G ]) Ab.id
    in ∫hom (h .fst) record { is-group-hom (h .snd) }

  .eta∘inv G      ext λ _  refl
  .inv∘eta G      ext λ _  refl
  .natural x y f  ext λ _  refl

For the other unitor, to give a map it suffices to give a map and we choose Again, that these are inverses follows by computation.

Ab-monoidal .unitor-r = to-natural-iso λ where
  .eta G  ∫hom  x  x , 1) record { pres-⋆ = λ x y  t-pres-*l }

  .inv G  R-adjunct (Tensor⊣Hom (Lift-ab _ ℤ-ab)) record where
    fst g = ∫hom  a  pow (Abelian→Group G) g (a .lower)) record
      { pres-⋆ = λ x y  pow-+ (Abelian→Group G) g (x .lower) (y .lower) }
    snd = record { pres-⋆ = λ x y  ext refl }

  .eta∘inv G      ext λ _  refl
  .inv∘eta G      ext λ _  refl
  .natural x y f  ext λ _  refl

Finally, the triangle and pentagon coherences are also trivial computations.

Ab-monoidal .associator = assc
Ab-monoidal .triangle = ext λ _ _      refl
Ab-monoidal .pentagon = ext λ _ _ _ _  refl