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 $n$”
 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:
Ex: Magmas
 Magma:

a set equipped with a binary operation $m: A^2 ⟶ A$
(no associativity, no unit)
 Commutative monoids:

m(x, y) = m(y, x)
Th: algebraic theory defines a monad $T: Set ⟶ Set$ in that way
and
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
Algebraic theory of the commutative rings
 $+, ×$ 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
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 $V = S = \lbrace true, false \rbrace$
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
 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 $h_1$:
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
Observation:
every object $A$ of the category $𝒞$ induces a $T$algebra $(TA, μ_A)$
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?
means that
In other words:
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