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Notes on Beck’s Distributive Laws
- L. Ze Wong
Notes on Becks Distributive Laws L. Ze Wong University of - - PowerPoint PPT Presentation
Notes on Becks Distributive Laws L. Ze Wong University of - - PowerPoint PPT Presentation
Notes on Becks Distributive Laws L. Ze Wong University of Washington, Seattle 2017 WARNING! The notation in this set of notes differs from Becks paper in the following key ways: Beck writes composites in the opposite direction: GF
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Motivation 1: Multiplication over Addition
Let S be the free monoid monad, T the free abelian group monad. ‘Multiplication distributes over addition’ means we have a map: STX → TSX e.g. (a + b)(c + d) → ac + ad + bc + bd where X = {a, b, c, . . . }, say. Further, TS is the free ring monad.
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Motivation 2: Tensoring monoids
Let A, B be monoids in a braided monoidal category (V, ⊗, 1). Then A ⊗ B is also a monoid, with multiplication A ⊗ B ⊗ A ⊗ B
A⊗tw⊗B
− − − − − → A ⊗ A ⊗ B ⊗ B
mA⊗mb
− − − − − → A ⊗ B where tw : B ⊗ A → A ⊗ B is given by the braiding.
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Monads in a 2-category
Fix a 2-category K. A monad in K consists of:
◮ 0-cell X ◮ 1-cell S : X → X ◮ 2-cells ηS : 1X ⇒ S and µS : SS ⇒ S
such that = = = i.e. a monad is a monoid in the monoidal category (End(X), ◦, 1X), for some 0-cell X.
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Distributive Law
A distributive law of S over T is a 2-cell ℓ : ST ⇒ TS such that: = ; = = =
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Characterization of Distributive Laws
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Characterization
Theorem (Beck 1969, Street 1972, Cheng 2011)
The following are equivalent:
- 1. Distributive laws ℓ : ST ⇒ TS,
- 2. Multiplications m : TSTS ⇒ TS s.t. (TS, ηTηS, m) is monad
satisfying the middle unitary law, and S
ηT S
= = ⇒ TS
TηS
⇐ = = T are monad morphisms.
- 3. Liftings of the monad T to a monad ˜
T over XS,
- 4. Extensions of the monad S to a monad ˜
S over XT,
- 5. Certain elements of Mnd (Mnd(K)).
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The composite monad
Given ℓ : ST ⇒ TS, define m : TSTS ⇒ TS to be To get back ℓ, do:
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The composite monad
The middle unitary law holds: = and TηS : T ⇒ TS is a monad morphism: = = Similarly, ηTS : S ⇒ TS is a monad morphism.
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Liftings and Extensions
A lift of T to the EM object XS is a monad ˜ T: XS XS X
US ˜ T TUS
+ compatibility equations An extension of S to the Kleisli object XT is a monad ˜ S: X XT XT
FT S FT ˜ S
+ compatibility equations Kleisli objects in K are EM objects in Kop, so proofs for liftings hold for extensions too, by duality.
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Liftings and Extensions
Universal property1 of XS:
- Functors ˜
G : Y → XS ∼ =
- Functors G : Y → X
with S-action σ : SG ⇒ G
- XS
Y X
US ˜ G G
Let Y = XS, G = TUS, ˜ T = ˜
- G. Need S-action STUS ⇒ TUS.
Given by distributive law and canonical action of S on US: XS US T S
1In fact, this is an equivalence of categories
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Liftings and Extensions
Conversely, a lifting ˜ T means we have invertible 2-cells: US ˜ T T with inverse Lets us define a distributive law: := This works for lifts over any adjunction that gives S!
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Monads in Mnd(K)
Let (X, T), (Y, T ′) be monads in K. A monad opfunctor (F, φ) : (X, T) → (Y, T ′) consists of F : X → Y and φ : FT ⇒ T ′F X Y T ′ T F such that = and =
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Monads in Mnd(K)
A monad functor transformation is a 2-cell σ : F ⇒ F ′ such that F ′ F σ = F ′ F These form a 2-category Mnd∗(K). When X = Y, T = T ′, if (F, φ) : (X, T) → (X, T) is a monad, then F is a monad on X and φ is a distributive law of F over T! i.e.2 Dist(K) ∼ = Mnd∗(Mnd∗(K)) Also, Mnd∗ is a monad!
2Can define morphisms between distributive laws such that this is true!
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Algebras over TS
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Actions of T, S and TS
From before, have monad morphisms3: T
TηS
= = ⇒ TS
ηT S
⇐ = = S TηS = ; ηTS = These induce T- and S-actions on UTS, via the action of TS: ; In some sense, any TS-action is ‘captured’ by these two actions!
3Monad opfunctors with F = 1X.
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Actions of T, S and TS
Combining T- and S-actions on UTS gives canonical action of TS: S T XTS = UTS TS Can then show that the S-action ‘distributes over’ the T-action: =
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Algebras over TS
Let ℓ be a distributive law of S over T. From the characterization theorem, we get monads TS on X, ˜ T on XS and ˜ S on XT.
Theorem (Beck 1969, Cheng 2011)
The category of algebras of TS coincides with that of ˜ T. XTS ∼ = (XS)
˜ T
Dually, the Kleisli category of TS coincides with that of ˜ S. XTS ∼ = (XT) ˜
S
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Algebras over TS
Construct Φ : XTS → (XS) ˜
T and inverse Φ−1 as lifts arising from
universal properties of XS, (XS) ˜
T, XTS:
(XS) ˜
T
XTS XS XTS X
U ˜
T
Φ Φ−1 UTS US UTS
To get Φ−1, need S-action on UTS and ˜ T-action on lift of UTS. To get Φ, need TS action on USU ˜
T.
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Algebras over TS
We already have T- and S-actions on UTS. S-action gives a lift UTS : XTS → XS of UTS. To get ˜ T-action on UTS, lift4 T-action on UTS : XTS X XS ⇒ ⇒
T UTS UTS ˜ T UTS
- UTS
US
So we have Φ−1 : XTS → (XS)
˜ T
4Need T-action to be an S-alg. morphism, but this follows from
distributivity of S-action over T-action.
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Algebras over TS
To get Φ : (XS) ˜
T → XTS, need TS-action on USU ˜ T.
Use canonical actions of S on US and ˜ T on U ˜
T:
US U ˜
T
So XTS ∼ = (XS) ˜
T, and in fact,
UTSF TS = TS = USU
˜ TF ˜ TF S
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Distributivity of Adjoints
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Distributivity of adjoints
A distributive law gives rise to a ‘distributive square’: XTS XS XT X
U′ U ˜
T
F ˜
T
US UT F S F T US F S UT F T U′ U ˜
T
F ˜
T
where U′ is induced by the T-action on UTS. If certain coequalizers exist, U′ has a left adjoint5.
5Think of U′ as ‘restriction of scalars’, and adjoint as ‘extension of scalars’
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Distributivity of adjoints
Both composites XTS → X are the same: USU ˜
T = UTU′.
Both composites XS → XT are the same: U′F ˜
T = F TUS.
XTS XS XT X
U′ U ˜
T
F ˜
T
US UT F S F T US F S UT F T U′ U ˜
T
F ˜
T
This is a distributive adjoint situation, and there is an adjunction: Dist Adj X (Dist X)op
Struc Sem
⊣
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Distributivity of adjoints
If U′ has an adjoint F ′: XTS XS XT X
U′ U ˜
T
F ˜
T
US UT F ′ F S F T u f e−1 e′
To get distributive law: Need isomorphisms u, f that are ‘dual’ to each other. These give rise to e, e′. But e goes in the ‘wrong’ direction, so need e to be an isomorphism too, to get e−1.
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Thank you!
Questions?
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