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 $\Sigma$:

is a family $({\Sigma }_n{)}_{n\in \mathbb{N}}$ of sets

An element $f\in {\Sigma }_n$ is called an “operation of arity $n$”

A term of a signature $\Sigma$ with variables in $X$:

is a finite tree of which nodes are the operations of $\Sigma$ and leaves are in $X\bigsqcup {\Sigma }_0$

NB: notation: $Term(\Sigma ,X)$

An algebraic theory of signature $\Sigma$:

is a family of pairs $(t,n)$ (also written $t=u$)

• $X=\{x_0,\cdots ,x_n\}$
• $t\in Term(\Sigma ,X)$
• $u\in Term(\Sigma ,X)$

Given a set $X$ of variables and algebraic theory:

$$T(X)=Term(\Sigma ,X)/\sim$$

where $\sim$ is the equivalence relation generated by the equations of the theory

Ex: The algebraic theory of monoids

• ${\Sigma }_2=\{m\}$

• ${\Sigma }_0=\{e\}$

• Signature:

  • ${\Sigma }_n=\varnothing$ for $n\ne 0,2$
  • $\Sigma =\{m:2,e:0\}$

• 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 $\Sigma$ with leaves in $TA$ (and also in ${\Sigma }_0$) modulo the equations of the algebraic theory.

For every such leaf $[t_i]\in TA$, one can pick an element $t_i\in Term(\Sigma ,A)$ such that $[A_i]$ is the equivalence class of $t_i$

Substitute $t_1,\cdots ,t_n$ in the tree $A\in Term(\Sigma ,TA)$ to obtain $t[t_1,\cdots ,t_n]\in Term(\Sigma ,A)$

The equivalence class $[\underset{\in Term(\Sigma ,A)}{\underbrace{t[t_1,\cdots ,t_n]}}]\in T(A)$

does not depend on the choice of $t_1,\cdots ,t_n$ but only on their equivalence classes

Algebraic theory of the commutative rings

• $+,\times$ are associative and commutative
• $0,1$: neutral elements
• $-$: inverse law
• distributivity

since we can push the $\times$ under the $+$ and $-$, by distributivity.

Algebras of algebraic theories

An algebra of an algebraic theory:

is a set $A$ equipped with a function $op\stackrel{\scriptscriptstyle\mathrm{def}}{=}f:A^n⟶A$ for every operation $f\in {\Sigma }_n$ 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 $h(o{p}_A(a_1,\cdots ,a_n))=o{p}_B(h(a_1),\cdots ,h(a_n))$ for all $op\in {\Sigma }_n,a_1,\cdots ,a_n\in A$

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 $A$ = $S⟹(S\times A)$

To make things simpler, let us take $V=S=\{true,false\}$

and

• $updat{e}_{⟨true⟩}:A⟶A$
• $updat{e}_{⟨false⟩}:A⟶A$

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 $TA=\underset{n\in \mathbb{N}}{⨆}{A}^n$?

• unit of the monad: ${\eta }_A:A⟶TA$
• multiplication of the monad: ${\mu }_A:TTA⟶TA$

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: $TAhA{\eta }_{A}i{d}_{A}A$

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: $TATf{h}_{A}TB{h}_{B}AfB$

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” $T:Set⟶Set$, $TA\stackrel{\scriptscriptstyle\mathrm{def}}{=}\underset{n\in \mathbb{N}}{⨆}{A}^n$

a function $h:TA⟶A$ is the same thing as a family of functions $h_n:A^n⟶A$

NB: $h_n(a_1,\cdots ,a_n)$ can be thought of as $a_1\cdots a_n$

In particular:

• neutral element: $h_0:A^0=\{\star \}⟶A$
• multiplication: $h_2:A^2⟶A$

Claim: $h_2(h_2(a,b),c)=h_3(a,b,c)$

Because of $[[a,b],c]\in TTATh{\mu }_{A}TA\ni [h_2(a,b),h_1(c)]h[a,b,c]\in TAhA\ni h_2(h_2(a,b),h_1(c))=h_3(a,b,c)$

Claim: $h_1(a)=a$

Because of $[a]\in TAha\in A{\eta }_{A}i{d}_{A}A\ni h_1(a)=a$

More generally:

$$h_2(h_p(a_1,\cdots ,a_p),h_q(a_1^{\prime },\cdots ,a_q^{\prime }))=h_{p+q}(a_1,\cdots ,a_p,a_1^{\prime },\cdots ,a_q^{\prime })$$

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 $(TA,{\mu }_A)$

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 ($A_{\infty }$-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: