Cours 6: Algebraic theories
Free monads
 Pointed set:

a set $A$ equipped with an element
One should think of $\ast$ as a $0$ary operation $1 = A^0 ⟶ A$.
 $n$ary operation:

element of $A^n ⟶ A$
Monoid:
 set equipped with binary operation $A^2 ⟶ A$
 with associativity law: $m(m(x, y), z) = m(x, m(y, z))$
Partiality monad: $1+A$ ⟺ the “free pointed set” monad
Algebraic theories
 A signature $Σ$:

is a family $(Σ_n)_{n∈ℕ}$ of sets
An element $f∈Σ_n$ is called an “operation of arity $n$”
\[f ∈ Σ_2 \quad f(f(x, y), f(f(x, y), y))\] A term of a signature $Σ$ with variables in $X$:

is a finite tree of which nodes are the operations of $Σ$ and leaves are in $X \sqcup Σ_0$
NB: notation: $Term(Σ, X)$
 An algebraic theory of signature $Σ$:

is a family of pairs $(t, n)$ (also written $t = u$)
 $X = \lbrace x_0, ⋯, x_n \rbrace$
 $t ∈ Term(Σ, X)$
 $u ∈ Term(Σ, X)$
 Given a set $X$ of variables and algebraic theory:
 \[T(X) = Term(Σ, X)/\sim\]
where $\sim$ is the equivalence relation generated by the equations of the theory
Ex: The algebraic theory of monoids
 $Σ_2 = \lbrace m \rbrace$

$Σ_0 = \lbrace e \rbrace$

Signature:
 $Σ_n = ∅$ for $n ≠ 0, 2$
 $Σ = \lbrace m:2, e:0 \rbrace$

Equations:
 associativity: $m(m(x, y), z)$
 unit laws: $m(e, x) = x = m(x, e)$
In the case of the algebraic theory of monoids:
\[T(X) = \left\lbrace \text{trees generated by m, e with leaves in X} \right\rbrace / \text{assoc, unit laws} \\ = \lbrace \text{finite words on } X \rbrace = \bigsqcup_{n∈ℕ} A^n\]Ex: Magmas
 Magma:

a set equipped with a binary operation $m: A^2 ⟶ A$
(no associativity, no unit)
\[T(X) = \lbrace \text{binary trees with leaves in } X\rbrace = Term(Σ, X)\] Commutative monoids:
 \[m(x, y) = m(y, x)\]
\[η: \begin{cases} A &⟶ TA \\ a &⟼ \text{the equivalence class of } a ∈ Term(Σ, A) \end{cases}\]Th: algebraic theory defines a monad $T: Set ⟶ Set$ in that way
and
\[μ: TTA ⟶ TA\]natural in $A$.
 An element of $T(TA)$:

is a tree of signature $Σ$ with leaves in $TA$ (and also in $Σ_0$) modulo the equations of the algebraic theory.
For every such leaf $[t_i] ∈ TA$, one can pick an element $t_i ∈ Term(Σ, A)$ such that $[A_i]$ is the equivalence class of $t_i$
Substitute $t_1, ⋯, t_n$ in the tree $A ∈ Term(Σ, TA)$ to obtain $t[t_1, ⋯, t_n] ∈ Term(Σ, A)$
The equivalence class \([\;\underbrace{t[t_1, ⋯ , t_n]}_{∈ Term(Σ,A)}\;] ∈ T(A)\)
does not depend on the choice of $t_1, ⋯, t_n$ but only on their equivalence classes
\[[t] ∈ TTA, \quad [t_i]∈ TA\]Algebraic theory of the commutative rings
\[Σ = \lbrace 0: 0, 1: 1, :1, ×: 2, +:2 \rbrace\] $+, ×$ are associative and commutative
 $0, 1$: neutral elements
 $$: inverse law
 distributivity
since we can push the $×$ under the $+$ and $$, by distributivity.
Algebras of algebraic theories
 An algebra of an algebraic theory:

is a set $A$ equipped with a function $op ≝ f: A^n ⟶ A$ for every operation $f ∈ Σ_n$ of arity $n$, and satisfies the equations of the theory.
 A homomorphism $A ⟶ B$ between such algebras:

is a function $h: A ⟶ B$ such that \(h(op_A(a_1, ⋯, a_n)) = op_B(h(a_1), ⋯, h(a_n))\) for all $op∈ Σ_n, \; a_1, ⋯, a_n ∈ A$
Th: $TA$ is the free algebra generated by the set $A$ in the sense that for every function $f: A ⟶ B$ where $B$ is an algebra, there exists a unique homomorphism \(h: TA ⟶ B\) such that
\[\begin{xy} \xymatrix{ TA \ar[r]^h & B \\ A \ar[u]^{η_A} \ar[ru]_f } \end{xy}\]commutes
The state monad
Fact: the state monad can be defined as the monad associated to a specific algebraic theory of mnemoids.
The free mnemoid generated by $A$ = $S ⟹ (S × A)$
To make things simpler, let us take $V = S = \lbrace true, false \rbrace$
\[lookup: \begin{cases} A^2 &⟶ A \\ (M, N) &⟼ \begin{cases} M \text{ if the register is } true \\ N \text{ else} \end{cases} \end{cases}\]and
 $update_{⟨true⟩}: A ⟶ A$
 $update_{⟨false⟩}: A ⟶ A$
How does one recover the algebra of a given alg. theory from the monad $T$?
For instance: can we recover the notion of monoid from the monad $TA = \bigsqcup_{n∈ℕ} A^n$?
 unit of the monad: \(η_A: A ⟶ TA\)
 multiplication of the monad: \(μ_A: TTA ⟶ TA\)
 An algebra of a monad $T: 𝒞 ⟶ 𝒞$:

is an object $A$ of the category $𝒞$ equipped with a morphism $h: TA ⟶ A$ s.t. the following diagrams commute: \(\begin{xy} \xymatrix{ TA \ar[rd]^h \\ A \ar[u]^{η_A} \ar[r]_{id_A} & A } \end{xy}\)
and
\[\begin{xy} \xymatrix{ TTA \ar[r]^{Th} \ar[d]_{μ_A} & TA \ar[d]^h \\ TA \ar[r]_h & A } \end{xy}\] A homomorphism between $T$algebras $f: (A, h_A) ⟶ (B, h_B)$:

is a morphism $f: A ⟶ B$ s.t. the following diagram commutes: \(\begin{xy} \xymatrix{ TA \ar[r]^{Tf} \ar[d]_{h_A} & TB \ar[d]^{h_B} \\ A \ar[r]_f & B } \end{xy}\)
This defines a category $Alg$ whose objects are the $T$algebras whose morphisms are the homomorphisms between them.
Th: for every algebraic theory, the category of algebras (and homomorphisms) of the theory is isomorphic to the category of $T$algebras for the associated monad $T$.
Example: in the case of the “free monoid monad” $T: Set ⟶ Set$, \(TA ≝ \bigsqcup_{n∈ℕ} A^n\)
a function $h: TA ⟶ A$ is the same thing as a family of functions \(h_n: A^n ⟶ A\)
NB: $h_n(a_1, ⋯, a_n)$ can be thought of as $a_1 ⋯ a_n$
In particular:
 neutral element: $h_0: A^0 = \lbrace \star \rbrace ⟶ A$
 multiplication: $h_2: A^2 ⟶ A$
Claim: $h_2(h_2(a, b), c) = h_3(a, b, c)$
Because of \(\begin{xy} \xymatrix{ [[a, b], c] ∈ TTA \ar[r]^{\qquad Th \qquad } \ar[d]_{μ_A} & TA\ni [h_2(a,b), h_1(c)] \ar[d]^h \\ [a, b, c] ∈ TA \ar[r]_{h \qquad \qquad} & A \ni h_2(h_2(a, b), h_1(c))=h_3(a, b, c) } \end{xy}\)
Claim: $h_1(a) = a$
Because of \(\begin{xy} \xymatrix{ [a] ∈ TA \ar[rd]^h \\ a ∈ A \ar[u]^{η_A} \ar[r]_{id_A} & A \ni h_1(a)=a } \end{xy}\)
More generally:
\[h_2(h_p(a_1, ⋯, a_p), h_q(a_1', ⋯, a_q')) = h_{p+q}(a_1, ⋯, a_p, a_1', ⋯, a_q')\]
Hence: a $T$algebra for the free monoid monad $T$ is entirely determined by $h_0$ and $h_1$:
\[h_n(a_1, ⋯, a_n) = h_2(a_1, h_2(a_2, ⋯, a_n))\]Associativity comes from the equation \(h_2(a, h_2(b, c)) = h_3(a, b, c) = h_2(h_2(a, b), c)\)
We can recover the monad from the notion of $T$algebra ⟶ the monad is an invariant of an algebraic theory.
From Monads to categories of $T$algebras
For every monad $T: 𝒞 ⟶ 𝒞$ , there is an adjunction
\[\xymatrix{ 𝒞 \ar@/^1cm/[r]^{F \text{ Free}} & Alg_T \ar@/^1cm/[l]^{𝒰 \text{ forget}} }\] \[𝒰: (A, h) ⟼ A\] \[F: A ⟼ (TA, \; μ_A: TTA ⟶ TA)\]Observation:
every object $A$ of the category $𝒞$ induces a $T$algebra $(TA, μ_A)$
\[μ_A: TTA ⟶ TA\]The monad $T = 𝒰 \circ F$ is recovered from the adjunction.
A monoid is a topological space equipped with a contractible space of operations ⟹ operads ($A_∞$algebras)
Operads: generalization of algebraic theories to a topological space of operations
Back to homorphisms: why are they so simple?
\[\begin{xy} \xymatrix{ \bigsqcup_{n∈ℕ} A^n = TA \ar[r]^{Tf} \ar[d]_{h_A} & TB = \bigsqcup_{n∈ℕ} B^n \ar[d]^{h_B} \\ A \ar[r]_f & B } \end{xy}\]means that
\[\begin{xy} \xymatrix{ A^n \ar[r]^{f^n} \ar[d]_{(h_A)_n} & B^n \ar[d]^{(h_B)_n} \\ A \ar[r]_f & B } \end{xy}\]In other words:
\[h_n(fa_1, ⋯, fa_n) = f(h_n(a_1, ⋯, a_n))\]
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