Lecture 6: Monoidal Categories

Teacher: Paul-André Melliès

\newcommand\xto\xrightarrow \newcommand\xfrom\xleftarrow

We have maps:

A \overset{Δ_A}{⟶} A × A\\ A \overset{!_A}{⟶} 1\\ A × B ⟶ A \qquad A × B ⟶ B

But we don’t have a non trivial clique in $A ⊸ A ⊗ A$

Recall that a monoidal category is a category equipped with a functor

⊗: 𝒞 × 𝒞 ⟶ 𝒞

and an object $I$, together with a family of isomorphisms

α_{A,B,C}: (A ⊗ B) ⊗ C ⟶ A ⊗ (B ⊗ C)\\ λ_A: I ⊗ A ⟶ A\\ ρ_A: A ⊗ I ⟶ A

natural in $A,B,C$


In a cartesian category:

\begin{xy} \xymatrix{ A & & 1 \\ & A × 1 \ar[lu]^{π_1 = λ_A} \ar[ru]_{π_2} &\\ & A \ar@{.>}[u]_{\bar {λ_a}} \ar@/^2pc/[luu]^{id_A} \ar@/_2pc/[ruu]_{!_A} & } \end{xy}

We can show that $\bar λ_A = λ_A^{-1}$:

λ_A \bar λ_A = id\\ \bar λ_A λ_A = id

Naturality for $α$:

\begin{xy} \xymatrix{ (A⊗ B)⊗C \ar[r]^{α_{A,B,C}} \ar[d]_{(h_A ⊗ h_B) ⊗ h_C} & A⊗(B⊗C) \ar[d]^{h_A ⊗ (h_B ⊗ h_C)} \\ (A'⊗B')⊗C' \ar[r]_{α_{A',B',C'}} & A'⊗(B'⊗C') } \end{xy}

If we stop here: not enough!

Coherence diagrams

Moreover, one requires that the diagrams below commute:

Here, the associativity morphism can depicted as:

  digraph {
    rankdir=TB;
    t1[label="⊗", shape=rectangle];
    t2[label="⊗", shape=rectangle];
    t1 -> C;
    t1 -> t2 -> A;
    t2 -> B;
  }

$\leadsto$

  digraph {
    rankdir=TB;
    t1[label="⊗", shape=rectangle];
    t2[label="⊗", shape=rectangle];
    t1 -> A;
    t1 -> t2 -> B;
    t2 -> C;
  }

MacLane’s pentagon is required to commute, so that there’s only one morphism going from one parenthesized expression to another equivalent (where associativity has been used) one.

\begin{xy} \xymatrix@C=5em{ ((A\otimes B)\otimes C)\otimes D\ar[d]_{\alpha_{A\otimes B,C,D}}\ar[r]^{\alpha_{A,B,C}\otimes {1_D}} &(A\otimes(B\otimes C))\otimes D\ar[r]^{\alpha_{A,B\otimes C,D}} &A\otimes((B\otimes C)\otimes D)\ar[d]^{ {1_A}\otimes\alpha_{B,C,D}}\\ (A\otimes B)\otimes(C\otimes D)\ar[rr]_{\alpha_{A,B,C\otimes D}}&&A\otimes(B\otimes (C\otimes D)) } \end{xy}

The MacLane pentagon enforces that for critical pairs, the fact that $⊗$ is a functor guarantees it for squares whith no critical pair.

Similarly: coherence diagram for $I$


Thm (Coherence theorem): There is a unique structural morphism from a binary tree of $⊗$’s into another binary tree of tensors.

cf. Stassheff associahedra / $A_∞$-algebras

NB: Discrete monoidal category = a monoid

Braided monoidal category = a commutative monoid when it is discrete

Symmetric braided monoidal category (cf. picture) : $γ_{A,B}^{-1} = γ_{A,B}$

Examples:

  • Monoidal: endofunctors of a category $𝒞$: $End(𝒞)$ (no hope to get a braiding, as $FG ≠ GF$ in general)

  • Braided: the category $Braid$, whose

    • objects are natural numbers
    • morphisms are braids (cf. picture). Equality between two morphisms is topological.
  • Symmetric: $Coh$ with tensor product $⊗$ and unit $1$

Strict monoidal category:

when $α, λ$ and $ρ$ are identities

γ_{A, B}: A ⊗ B ⟶ B ⊗ A

(cf. Reidemeister moves)

  • Opposite category of a cartesian product = category with coproducts (not cartesian anymore!)
  • But: opposite of a symmetric monoidal category = remains a symmetric monoidal category!

In $Coh$:

A \xto {f} A' \qquad B \xto {g} B'

then define

A ⊗ B \xto {f ⊗ g} A' ⊗ B'

by

f ⊗ g ≝ \lbrace((a,b), (a',b')) \; \mid \; (a,a') ∈ f \text{ and } (b, b') ∈ f\rbrace

Exercise: show that $f ⊗ g$ just defined is a clique of the coherence space $(A ⊗ B) ⊸ (A’ ⊗ B’)$

Sometimes, it is convenient to write

  • $a ⊗ b$ for the element $(a,b) ∈ \vert A ⊗ B \vert$
  • $a ⊸ b$ for the element $(a,b) ∈ \vert A ⊸ B \vert$

for $a ∈ \vert A \vert, \; b ∈ \vert B \vert$.

So:

f ⊗ g ≝ \lbrace (a ⊗ b) ⊸ (a' ⊗ b') \; \mid \; a ⊸ a' ∈ f \text{ and } b ⊸ b' ∈ g\rbrace

In $Coh$:

γ_{A,B}: A ⊗ B ⟶ B ⊗ A

is defined as the clique of $(A ⊗ B) ⊸ (B ⊗ A)$:

γ_{A,B} ≝ \lbrace (a ⊗ b) ⊸ (b ⊗ a) \; \mid \; a ∈ \vert A \vert \text{ and } b ∈ \vert B \vert\rbrace

Goal: understand the structure of the category of coherence spaces

Symmetric monoidal closed category

Same as symmetric cartesian closed category, except that

  • $×$ is replaced by $⊗$
  • $⇒$ is replaced by $⊸$
A ⊗ \bullet ⊣ A ⊸ \bullet: 𝒞 ⟶ 𝒞

Multiplicative Intuitionistic Linear Logic

Sequent Calculus VS ND:

  • in ND: intro/elim rules
  • in SC: only intro rules, but you also have a cut (which eliminates formulas)
A, B ≝ 1 \; \mid \; A ⊗ B \; \mid \; A ⊸ B \; \mid \; α
  • axiom

    \cfrac{}{A ⊢ A}
  • $⊸$ right/left

    \cfrac{Γ, A ⊢ B}{Γ ⊢ A ⊸ B} \qquad \cfrac{Δ ⊢ A \qquad Γ, B ⊢ C}{Γ, Δ, A ⊸ B ⊢ C}
  • $⊗$ right/left

    \cfrac{Γ ⊢ A \qquad Δ ⊢ B}{Γ, Δ ⊢ A ⊗ B} \qquad \cfrac{Γ, A, B ⊢ C}{Γ, A ⊗ B ⊢ C}
  • $1$ left/right \cfrac{Γ ⊢ A}{Γ, 1 ⊢ A} \qquad \cfrac{}{⊢ 1}
  • cut

    \cfrac{Δ ⊢ A \qquad Γ, A ⊢ B}{Γ, Δ ⊢ B}
  • exchange

    \cfrac{Γ, A_1, A_2, Δ ⊢ B}{Γ, A_2, A_1, Δ ⊢ B}

Suppose given a SMC (symmetric monoidal closed) category $𝒞$ together with an object $⟦α⟧ ∈ 𝒞$ for every type variable.

Every derivation tree $π$ of a sequent

A_1, ⋯, A_n ⊢ B

is a morphism interpreted as

⟦A_1⟧ ⊗ ⋯ ⊗ ⟦A_n⟧ \xto {⟦π⟧} ⟦B⟧
⟦A \overbrace{⊗}^{\text{syntax}} B⟧ \, ≝ \, ⟦A⟧ \overbrace{⊗}^{\text{semantics}} ⟦B⟧\\ ⟦A ⊸ B⟧ \, ≝ \, ⟦A⟧ ⊸ ⟦B⟧

Axiom: $A \xto {id_A} A$, etc…

Moreover: interpretation of derivation trees is invariant modulo cut elimination.

π_1 \overset{\text{cut-elim}}{⟶} π_2 ⟹ ⟦π_1⟧ = ⟦π_2⟧

$Coh$ is SMC, so it can provide an interpretation of MLL.

$\star$-autonomous categories

In a SMCC, with $eval$, you have a map:

(A ⊸ ⊥) ⊗ A ⟶ ⊥

which can be turned into

A ⟶ (A ⊸ ⊥) ⊸ ⊥

Prop: every $\star$-autonomous category comes equipped with a symmetric monoidal structure defined as (De Morgan duality):

A ⅋ B \, ≝ \, (A ⊸ ⊥ ⊗ B ⊸ ⊥) ⊸ ⊥

where $⊥$ is the unit.

Important here: $⅋$ is associative.

\bar α_{A,B,C}: \begin{cases} (A ⅋ B) ⅋ C &⟶ A ⅋ (B ⅋ C) \\ A ⅋ ⊥ &\overset{\bar ρ_A}{⟼} A\\ ⊥ ⅋ A &\overset{\bar λ_A}{⟼} A \end{cases}

Given a $\star$-autonomous category $𝒞$ and an interpretation $⟦α⟧ ∈ 𝒞$ of every type variable $α$, we associate to every derivation tree $π$ of the sequent $⊢ A_1, …, A_n$:

1 \overset{⟦π⟧}{⟶} ⟦A_1⟧ ⅋ ⋯ ⅋ ⟦A_n⟧

or:

A_1^⊥ ⊗ ⋯ ⊗ A_n^⊥ ⟶ ⊥ \text{ where } A^⊥ ≝ A ⊸ ⊥

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