Paolo Capriotti's blog https://www.paolocapriotti.com/atom.xml Paolo Capriotti paolo@capriotti.io 2019-04-30T00:00:00Z O(2) as a semidirect product https://www.paolocapriotti.com/blog/2019/04/30/o2-as-a-semidirect-product/index.html 2019-04-30T00:00:00Z 2019-04-30T00:00:00Z It is a simple linear algebra exercise to show that $$O(n)$$, the group of orthogonal $$n × n$$ real matrices, is the semidirect product of its subgroup $$SO(n)$$, consisting of matrices of determinant 1, and the cyclic group $$C_2$$ on two elements.

In other words, there is a short exact sequence of Lie groups $1 → SO(n) → O(n) → O(1) → 1,$ where the map $$O(n) → O(1)$$ is the determinant, and this sequence splits through any homomorphism $$O(1) → O(n)$$ that maps $$-1$$ to a reflection.

We can replicate this construction in HoTT, but unfortunately only for the case $$n = 2$$, where, thanks to some fortuitous coincidences, we can obtain (the homotopy types of) the Lie groups above and their deloopings.

First, let $$\mathsf{Aut}(F)$$ denote the automorphism group of a type $$F$$, i.e. the type $$F ≅ F$$. If $$F : \mathcal{U}$$, where $$\mathcal{U}$$ is a univalent universe, then we can define a delooping of $$\mathsf{Aut}(F)$$ simply as the connected component of $$F$$ in $$\mathcal{U}$$, i.e. $B\mathsf{Aut}(F) = (X : \mathcal{U}) × \| X = F \|.$

Now we can define: $\begin{array}{l} O(1) :≡ \mathsf{Aut}(\underline{2}), \\ O(2) :≡ \mathsf{Aut}(S¹). \\ \end{array}$

The first definition is justified by the fact that $$O(1)$$ is the discrete group with two elements, which is isomorphic to the permutation group $$Σ_2 ≡ \mathsf{Aut}(\underline{2})$$.

As for the second, consider the function $$(S^1 → S^1) → S^1$$ that maps a function to its value on the base point. One can show that the fibres of this map form the constant $$ℤ$$ family. So $$(S^1 → S^1) ≅ S^1 × ℤ$$, where composition acts like multiplication on the second component. It follows that $$\mathsf{Aut}(S^1) ≅ S^1 × ℤ^* ≅ S^1 × \mathsf{Aut}(\underline{2})$$. This argument shows that the inclusion $$O(2) → \mathsf{Aut}(S^1)$$ is a homotopy equivalence in spaces, which justifies our definition.

To replicate the exact sequence above in type theory, we have to construct it at the level of deloopings. Let $$p: \mathcal{U}→ \mathcal{U}$$ be defined by $p(X) :≡ \| X = S¹ \|_0.$ It follows from the above calculations of $$\mathsf{Aut}(S¹)$$ that $$p(S¹) = \underline{2}$$. Therefore, $$p$$ maps the connected component of $$S¹$$ to the connected component of $$\underline{2}$$, hence it defines a function $$p: BO(2) → BO(1)$$. Define $$BSO(2)$$ to be the fibre of $$p$$.

Lemma 1. For $$A : BO(1)$$, there is an equivalence $$(\underline{2} = A) ≅ A$$.

Proof. Let $$ϕ_A : (\underline{2} = A) → A$$ be defined by $$ϕ_A(α) = α(0)$$, where we are implicitly regarding an element of $$\underline{2} = A$$ as an equivalence. Since $$ϕ_{\underline{2}}$$ is obviously an equivalence, it follows that $$ϕ_A$$ is an equivalence for all $$A$$ in the connected component of $$\underline{2}$$.$$\square$$

Proposition 2. The map $$\| BO(2) \|_1 → BO(1)$$ induced by $$p$$ is an equivalence.

Proof. Since both types are connected, it is enough to show that the induced map on the loop spaces is an equivalence. For any $$X : BO(2)$$, let $$ϕ: (X = S^1) → p(X)$$ be the map obtained by composing $$\mathsf{ap}_p: (X = S^1) → (p(X) = p(S¹))$$ with the equivalence $$(p(X) = p(S¹)) → p(X)$$ given by Lemma 1, and using the fact that $$p(S¹) = \underline{2}$$. We will show that $$ϕ$$ is equal to the canonical projection into the 0-truncation.

By path elimination, it is enough to show that the reflexivity path $$S^1 = S^1$$ is mapped to its image in the 0-truncation, which follows immediately from expanding the definitions.

In particular, the map $$p(X) → p(X)$$ induced by $$ϕ$$ is an equivalence, hence $$\mathsf{ap}_p$$ is an equivalence by 2-out-of-3.$$\square$$

Corollary 3. Ω(BSO(2)) ≅ S¹

Proof. By identifying $$Ω\| BO(2) \|$$ with $$ΩBO(1) = \underline{2}$$, we get that the map induced by $$p$$ on the loop spaces is equivalent to the second projection $$S¹ × \underline{2} → \underline{2}$$. In particular, its fibre is $$S¹$$. The conclusion then follows from the long exact sequence of homotopy groups.$$\square$$

So we have the desired fibre sequence $$BSO(2) → BO(2) → BO(1)$$. To show that our definition of $$BSO(2)$$ matches what we have in spaces, there is one more subtle point that we need to address. We only showed that the loop space of $$BSO(2)$$ is equivalent to $$S¹$$ as a space, but a priori there could be another ∞-group structure on $$S¹$$, corresponding to another delooping. Fortunately, though, $$S¹$$ is the Eilenberg-MacLane space $$K(ℤ, 1)$$, hence any delooping of it must be a $$K(ℤ, 2)$$, which means that there is exactly one. In particular, $$BSO(2)$$ as we defined it is a possible model for the infinite dimensional complex projective space, another one being what arises from the type-theory construction of Eilenberg-MacLane spaces, namely $$\| S² \|_2$$.

The fibre sequence $$BSO(2) → BO(2) → BO(1)$$ gives rise to an exact sequence of $$∞$$-groups $$SO(2) → O(2) → O(1)$$. Such a sequence splits if and only if the second map has a section.

Since the suspension of $$\underline{2}$$ is equivalent to $$S^1$$, suspension can be regarded as a pointed map $$Σ : BO(1) → BO(2)$$. As usual, we will denote by $$\mathsf{merid}: A → N = S$$ the path constructor in the definition of the suspension $$Σ A$$, and take the north pole $$N$$ as the base point of $$Σ A$$.

Proposition 4. The suspension map $$Σ: BO(1) → BO(2)$$ is a section of $$p: BO(2) → BO(1)$$.

Proof. For any type $$A$$ in $$BO(1)$$, let $$σ_A : A → A$$ be the only non-identity equivalence of $$A$$, and define a function $$ψ_A : A → S^1 → ΣA$$ so that $$ψ_A(a)$$ corresponds to the loop $$\mathsf{merid}(a) · \mathsf{merid}(σ_A(a))^{-1}$$.

Then for all $$A : BO(1)$$, and all $$a: A$$, $$ψ_A(a)$$ is an equivalence, since this is true for $$A ≡ \underline{2}$$, and being an equivalence is a proposition. We have thus defined a map $$A → (S¹ ≅ ΣA)$$, hence in particular a map $$A → \| S^1 ≅ ΣA \|_0$$. It is easy to verify that the latter map is an equivalence for $$A ≡ 2$$, which implies that it is an equivalence for all $$A : BO(1)$$. This shows that $$Σ$$ is a section of $$p$$, concluding the proof.$$\square$$

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Deloopings of 1-groups https://www.paolocapriotti.com/blog/2019/01/02/classifying-spaces/index.html 2019-01-02T00:00:00Z 2019-01-02T00:00:00Z By a delooping of a space $$X$$ we mean a connected pointed space $$Y$$ such that $$X ≅ ΩY$$. Using higher inductive types in HoTT, one can construct deloopings of 1-groups, i.e. sets with a group structure. This follows from the more general construction of Eilenberg-MacLane spaces (Licata and Finster 2014).

Based on the fact that 1-groups can be delooped, one can define an $$∞$$-group to be a space that can be delooped. Classically, there are more concrete definitions of $$∞$$-groups, often as algebras of certain operads, but they are not directly replicable in HoTT. Examples of $$∞$$-groups that are not $$1$$-groups include matrix Lie Groups such as $$O(n)$$, $$SO(n)$$, $$U(n)$$ and $$SU(n)$$. Neither these groups themselves, nor their deloopings, have been constructed in HoTT, except for a few special cases (for example, $$U(1) = S^1$$, $$SO(3) = ℝP^3$$ and $$SU(2) = S^3$$).

Furthermore, there is no general technique that allows us to construct deloopings of such objects. In a way, this is to be expected, since we don’t yet know how to express their full $$∞$$-group structure in HoTT in way that is not simply exhibiting a delooping. Therefore, it is not yet known whether deloopings of $$S^3$$ or $$ℝP^3$$ exist in HoTT.

However, some special cases can be dealt with, for example, automorphism groups $$\mathsf{Aut}(X)$$, for a type $$X$$ in a universe $$\mathcal{U}$$, defined as $$\mathsf{Aut}(X) :≡ (X =_{\mathcal{U}} X)$$. In fact, if we define $$B\mathsf{Aut}(X)$$ to be the image of the map $$1 → \mathcal{U}$$ determined by $$X$$, it is easy to see that $$ΩB\mathsf{Aut}(X) ≅ (X =_{\mathcal{U}} X) ≡ \mathsf{Aut}(X)$$. These kinds of definitions are quite convenient, since they are only based on the existence of a univalent universe containing $$X$$, and propositional truncation (needed to define images of maps). In particular, they do not require the existence of any other higher inductive type besides propositional truncation.

One downside is that this construction of $$B\mathsf{Aut}(X)$$ raises the universe level, since it ends up being as large as the universe $$\mathcal{U}$$, hence larger than $$X$$ and $$\mathsf{Aut}(X)$$. Furthermore, clearly not all $$∞$$-groups are of the form $$\mathsf{Aut}(X)$$ for some space $$X$$. On the other hand, we are not limited to considering only bare types $$X$$, but we can look at automorphism groups in other categories. Using this observation, we can actually replicate the construction of Eilenberg-MacLane spaces of the form $$K(G,1)$$ only using a univalent universe of sets and propositional truncation. The idea is to express any $$1$$-group $$G$$ as the group of automorphism of an object in a univalent category, and then use the same construction as in the case of $$B\mathsf{Aut}(X)$$.

Let $$G$$ be a 1-group, i.e. a set with a group structure. A $$G$$-set is defined to be a set $$X$$, together with a group homomorphism $${G}^{\mathrm{op}} → (X ≅ X)$$. If $$g : G$$ and $$x : X$$, let $$x · g$$ denote the result of the action of $$g$$ on $$x$$.

If $$X$$ and $$Y$$ are $$G$$-sets, and $$f : X → Y$$ is a function, we say that $$f$$ is equivariant if for all $$x : X$$ and $$g : G$$, we have that $$f(x) · g = f(x · g)$$.

More abstractly, we can regard $$G$$ as a category with one object, denoted $$\mathbf{B}G$$, and define a $$G$$-set to be a presheaf on $$\mathbf{B}G$$. Then an equivariant function is simply a natural transformation, and $$G$$-sets with equivariant functions form a univalent category $$G\text{-}\mathsf{Set}$$. Note, however, that $$\mathbf{B}G$$ itself is not univalent, unless $$G$$ is the trivial group.

Clearly, $$G$$ itself can be made into a $$G$$-set via its right multiplication action. We will use the notation $$\underline G$$ to refer to $$G$$ regarded as a $$G$$-set. Note that $$\underline G$$ can be thought of as the representable presheaf corresponding to the unique object of $$\mathbf{B}G$$. It then follows from the Yoneda lemma that there is an equivalence $$G ≅ G\text{-}\mathsf{Set}(\underline G, \underline G)$$. Since $$\mathbf{B}G$$ is a groupoid, the monoid of endomorphisms of $$\underline G$$ is a group, which together with the previously stated equivalence and univalence of $$G\text{-}\mathsf{Set}$$ implies that $$G ≅ (\underline G = \underline G)$$.

Now let $$BG$$ denote the image of the map $$1 → G\text{-}\mathsf{Set}$$ corresponding to $$\underline G$$. In other words, we factor the map $$1 → G\text{-}\mathsf{Set}$$ into a $$(-1)$$-connected map $$b : 1 → BG$$ followed by a $$(-1)$$-truncated map $$i : BG → G\text{-}\mathsf{Set}$$. In particular, $$BG$$ is pointed and connected, and we will denote its basepoint also by $$b$$.

Now we can calculate: \begin{aligned} & G ≅ \\ & (\underline G = \underline G) ≅ \\ & (i(b) = i(b)) ≅ \\ & (b = b) ≅ \\ & \Omega B G, \end{aligned}

which shows that $$BG$$ is a delooping of $$G$$.

I formalised the above construction of $$BG$$ in agda. You can find it here.

Licata, Daniel R., and Eric Finster. 2014. “Eilenberg-MacLane Spaces in Homotopy Type Theory.” In Proceedings of the joint meeting of the twenty-third EACSL annual conference on Computer Science Logic (CSL) and the twenty-ninth Annual ACM/IEEE Symposium on Logic in Computer Science (LICS), 66:1–66:9. CSL-Lics ’14. ACM. https://doi.org/10.1145/2603088.2603153.

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Monads as lax functors https://www.paolocapriotti.com/blog/2015/06/22/monads-as-lax-functors/index.html 2015-06-22T00:00:00Z 2015-06-22T00:00:00Z Last week at FP Lunch I talked about how to generalise the notion of monad using lax functors to $$\mathsf{Cat}$$.

Let’s begin by reviewing the classical definition. A monad is given by the following data:

• a category $$\mathcal{C}$$;
• an endofunctor $$T : \mathcal{C}\to \mathcal{C}$$;
• natural transformations $$\eta : I \to T$$ and $$\mu : T \circ T \to T$$;

satisfying certains laws (namely: $$\mu \circ \eta T = \mu \circ T \eta = \mathsf{id}$$ and $$\mu \circ T \mu = \mu \circ \mu T$$). Note that the category $$\mathcal{C}$$ is considered to be part of the data, rather than fixed beforehand.

In this post, I will illustrate a compact formulation of the above definition that can easily be generalised to include other similar notions, which appear from time to time in functional programming.

Here is the punchline:

A monad is a lax 2-functor from the terminal 2-category 1 to $$\mathsf{Cat}$$.

To make sense of this definition, we need to venture into the marvellous world of higher categories. If we take the definition of category that we are familiar with, we can regard it as some sort of 1-dimensional structure: we have a set of objects, which we can picture as points, and a set of arrows between them, which we imagine as (oriented) 1-dimensional lines.

It is then relatively easy to go one dimension up, and imagine an entity with 3 levels of structure: objects, morphisms, and 2-dimensional “cells” connecting arrows. This is what we call a 2-category.

More precisely, a 2-category is given by:

• a set of objects (or 0-cells)
• for any two objects $$x, y$$, a set of morphisms (or 1-cells) $$\mathsf{hom}(x, y)$$;
• for any two objects $$x, y$$, and any two morphisms $$f, g : \mathsf{hom}(x, y)$$, a set of 2-morphisms $$\mathsf{hom}_2 (f, g)$$.

Of course, this cannot really be the complete definition of 2-category, as we also need to be able to compose morphisms and 2-morphisms, but we won’t go into much detail here. The interested reader can find more details on this nLab page.

The primary example of 2-category is $$\mathsf{Cat}$$, the 2-category of categories. Objects of $$\mathsf{Cat}$$ are (ordinary) categories (also called 1-categories), morphisms are functors, and 2-morphisms are natural transformations. Another example is the terminal 2-category, containing only 1 object, and no non-identity morphisms or 2-morphisms.

As always happens in mathematics, every new structure that we define is accompanied by a corresponding notion of morphism. Given 2-categories $$\mathcal{C}$$ and $$\mathcal{D}$$, we want to define what it means to give a “map” $$\mathcal{C}\to \mathcal{D}$$ that respects the 2-category structure. We call such maps 2-functors.

As it turns out, there are multiple ways to give a definition of 2-functor. They differ in the amount of strictness that they require. More precisely, a 2-functor $$\mathcal{C}\to \mathcal{D}$$ is given by:

• a function $$F$$ mapping objects of $$\mathcal{C}$$ to objects of $$\mathcal{D}$$;
• for any two objects $$x, y$$ of $$\mathcal{C}$$, a function (also denoted $$F$$) mapping morphisms between $$x$$ and $$y$$ in $$\mathcal{C}$$ to morphisms between $$F x$$ and $$F y$$ in $$\mathcal{D}$$;
• for any two objects $$x, y$$ of $$\mathcal{C}$$, and morphisms $$f, g : \mathsf{hom}(x, y)$$, a function mapping 2-morphisms between $$f$$ and $$g$$ to 2-morphisms between $$F f$$ and $$F g$$;

subject to certain “functoriality” properties. We can make this functoriality requirement precise in a number of different (non-equivalent) ways.

First, we might directly generalise the functoriality properties for functors, and require: \begin{aligned} & F\mathsf{id}= \mathsf{id}, \\ & F (g \circ f) = F g \circ F f. \end{aligned}

If we do that, we get the notion of strict functor. However, the elements appearing in the above equations are objects of certain categories (namely, $$\mathsf{hom}$$-categories of $$\mathcal{D}$$), and if category theory has taught us anything, it is the idea that comparing objects of categories up to equality is often not very fruitful.

Therefore, we are naturally lead to the notion of pseudofunctor, which weakens the equalities to isomorphisms: \begin{aligned} & F\mathsf{id}\cong \mathsf{id}, \\ & F (g \circ f) \cong F g \circ F f. \end{aligned}

However, we are interested in an even weaker notion here, called lax 2-functor, which replaces the isomorphisms above with arbitrary (possibly not invertible) 2-morphisms: \begin{aligned} & \mathsf{id}\to F \mathsf{id}, \\ & F g \circ F f \to F (g \circ f). \end{aligned}

The direction of the arrows can be reversed, yielding the dual notion of oplax functor, which we won’t need here.

Now we understand all the terminology used in the definition above. Let $$F : 1 \to \mathsf{Cat}$$ be a lax 2-functor. At the level of objects, $$F$$ maps the unique object of $$1$$ to $$\mathsf{Cat}$$, which amounts to just picking a single category $$\mathcal{C}$$. At the level of morphisms, we map the single (identity) morphisms of 1 to a functor $$T: \mathcal{C}\to \mathcal{C}$$. Now, the “lax structure” produces 2-morphisms in $$\mathsf{Cat}$$ (i.e. natural transformations): $$\eta : I \to T$$ and $$\mu : T \circ T \to T$$.

So it looks like lax 2-functors to $$\mathsf{Cat}$$, at least ignoring certain details that we haven’t discussed, correspond perfectly to the classical definition of monad. I encourage the interested reader to look at the complete definition of lax functor, and verify that everything does indeed match, including the monad laws.

After all this work, generalising the definition is now extremely easy: just replace the 2-category 1 with a more general category. A simple example is: given a monoid $$S$$, regard $$S$$ as a 2-category with 1 object and no non-trivial 2-morphism. Lax functors $$S \to \mathsf{Cat}$$ are exactly Wadler’s indexed monads.

It is also possible (although slightly more involved) to recover Atkey’s parameterised monads as lax functors. I’ll leave this as a fun exercise.

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Free monads in category theory (part 3) https://www.paolocapriotti.com/blog/2013/12/16/free-monads-part-3/index.html 2013-12-16T00:00:00Z 2013-12-16T00:00:00Z Introduction

In the previous post, we investigated free monads, i.e. those whose monad algebras are the same as algebras of some functor. In general, however, not all monads are free, not even in Haskell! Nevertheless, monad algebras can often be regarded as algebras of some functor, satisfying certain “algebraic laws”.

In the first post of the series, we looked at the list monad $$L$$. We observed that monad algebras of $$L$$ can be regarded as monoids, which is to say they are algebras of the functor $$F$$ given by $$F X = 1 + X²$$, subject to unit and associativity laws.

The list example is interesting, because it suggests a strong connection between monads and algebraic structures. Can we always regard algebraic structures (such as groups, rings, vector spaces, etc…) as the algebras of some monad?

In this post, we will try to generalise this example to other monads by developing a categorical definition of algebraic theory based on monads and monad algebras.

## Algebraic theories

The theory of monoids is a particular instance of a general pattern that occurs over and over in mathematics. We have a set of operations, each with a specified arity, and a set of laws that these operations are required to satisfy. The laws all have the form of equations with universally quantified variables.

For monoids, we have two operations: a unit $$e$$, which is a nullary operation (i.e. a constant), and multiplication $$·$$, a binary operation (written infix). The laws should be very familiar: \begin{aligned} e · x & = x \\ x · e & = x \\ x · (y · z) & = (x · y) · z \end{aligned} where every free variable is implicitly considered to be universally quantified.

As we observed in the first post of this series, the functor $$F$$ corresponding to the algebraic theory of monoids is given by $$F X = 1 + X²$$. Algebras of $$F$$ are sets equipped with the operations of a monoid, but there is no requirement that they satisfy the laws.

Since $$F$$ is polynomial, it has an algebraically free monad $$F^*$$, so $$F^* X$$ is in particular an $$F$$-algebra. If we focus on the first law above, we see that it just consists of a pair of terms in $$F^* X$$, parameterised over some unspecified element $$x : X$$. This can be expressed as a natural transformation: $X → F^*X × F^* X$ The same holds for the second law, while the third can be regarded as a function: $X³ → F^*X × F^*X$

We can assemble those three functions into a single datum, consisting of a pair of natural transformations: $X + X + X³ ⇉ F^* X$

If we set $$G X = X + X + X³$$, we have that the laws can be summed up concisely by giving a pair of natural transformations: $G ⇉ F^*,$ which, since algebraically free monads are free, is the same as a parallel pair of monad morphisms: $l, r : G^* ⇉ F^*,$ and this is something that we can easily generalise. Namely, we say that an algebraic theory is a parallel pair of morphisms of algebraically free monads.

Note that the terminology here is a bit fuzzy. Some authors might refer to the parallel pair above as a presentation of an algebraic theory. It ultimately depends on whether or not you want to consider theories with different syntactical presentations but identical models to be equal. With our definition, they would be considered different.

## Models

To really motivate this definition, we now need to explain what the models of an algebraic theory are. This is quite easy if we just follow our derivation of the general definition from the example.

We know that a monoid is an $$F$$-algebra $$θ : F X → X$$ that satisfies the monoid laws. Since $$F$$-algebras are the same as $$F^*$$-algebras, we can work with the corresponding $$F^*$$-algebra instead, which we denote by $$θ^* : F^*X → X$$.

This algebra satisfies the laws exactly when the two natural transformations above become equal when composed with $$θ^*$$, i.e. when $$θ^* ∘ l = θ^* ∘ r$$.

We thus define the category of models of an algebraic theory $$l, r : G^* ⇉ F^*$$ as the full subcategory of $$\mathsf{Alg}_F ≅ \mathsf{mAlg}_{F^*}$$ consisting of all those monad algebras $$θ^* : F^* X → X$$ such that $$θ^* ∘ l = θ^* ∘ r$$.

Now, we know that, in the case of monoids, this subcategory is monadic over $$\mathsf{Set}$$, but is this true in general?

We begin by defining the notion of a free model for some algebraic theory. In the monoid example, we used the list monad to build a monoid out of any set, and then proceeded to prove that this construction gives the left adjoint of the forgetful functor $$\mathsf{Alg}_F → \mathsf{Set}$$. This is of course the first step towards proving monadicity.

In general, there does not seem to be a way to generalise this construction directly. We pulled the list monad out of a hat, and showed that it was exactly the monad that we were looking for. We did not derive it using the functor $$F$$ in a systematic way that we could replicate in the general case.

Fortunately, there is another way to produce the free monoid over a set $$X$$. We start with the free $$F$$-algebra $$F^* X$$, then quotient it according to the laws. Intuitively, we define an equivalence relation that relates two elements $$t₁$$ and $$t₂$$ whenever there is a law that requires them to be equal.

The straightforward way to formalise this intuition is to take the equivalence relation generated by such pairs $$(t₁, t₂)$$, then take the corresponding quotient. A more conceptual approach is to say that $$T X$$ is obtained as a coequaliser: $G^* X ⇉ F^* X → T X.$

In the monoid example, $$F^* X$$ is the set of all trees with leaves labelled by elements of $$X$$. If we regard a tree as a parenthesised string of elements of $$X$$, then the equivalence relation on $$F^*$$ given by the coequaliser above corresponds to identifying strings with the same underlying list of elements but possibly different parenthesizations. Therefore, $$T X$$ is clearly isomorphic to the list monad.

## Conclusion

More generally, we can take any algebraic theory, which we defined as a parallel pair of monad morphisms between free monads $$F^*$$ and $$G^*$$, and take the coequaliser in the category of (finitary) monads.

With some reasonable assumptions on the functors $$F$$ and $$G$$, we can show that this coequaliser always exists, and that the algebras of the resulting monad are exactly the models of the algebraic theory we started with.

## Further reading

This concludes my series on the underlying theory of free monads and their relation with universal algebra.

Here is a list of resources where you can learn more about this topic:

• Michael Barr and Charles Wells, Toposes, Triples and Theories

“Triple” is the old term for monads. Chapter 3 is about the monadicity theorem and some of the material that I covered in this series.

• Saunders Mac Lane, Categories for the Working Mathematician

Chapter 6 is about monads and their algebras.

• Steve Awodey, Category Theory

The last chapter explains the relationship between initial algebras and monadic functors.

• Francis Borceux, Handbook of Categorical Algebra

A very comprehensive resource, with detailed proofs.

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Free monads in category theory (part 2) https://www.paolocapriotti.com/blog/2013/12/04/free-monads-part-2/index.html 2013-12-04T00:00:00Z 2013-12-04T00:00:00Z Introduction

In the previous post, I introduced the notion of monadic functor, exemplified by the forgetful functor from the category of monoids to $$\mathsf{Set}$$. We saw that monoids form a subcategory of the category of algebras of the functor $$F$$ defined by $$F X = 1 + X²$$, and we observed that those are the same as the monad algebras of the list monad.

## Algebraically free monads

More generally, we can try different subcategories of $$\mathsf{Alg}_F$$ and check whether they are monadic as well. So let’s start with possibly the simplest one: the whole of $$\mathsf{Alg}_F$$.

This leads us to the following definition: we say that an endofunctor $$F$$ admits an algebraically free monad if $$\mathsf{Alg}_F$$ is monadic. The corresponding monad is called the algebraically free monad over $$F$$.

Informally, the algebraically free monad over $$F$$ is a monad $$T$$ such that monad algebras of $$T$$ are the same as functor algebras of $$F$$.

Unfortunately, not all functors admit an algebraically free monad. For example, it is easy to see that the powerset functor does not.

## Free monads as initial algebras

The free package on Hackage defines something called “free monad” for every Haskell functor. What does this have to do with the notion of algebraically free monad defined above?

Here is the definition of Free from the above package:

data Free f a
= Pure a
| Free (f (Free f a))


Translating into categorical language, we can define, for an endofunctor $$F$$, the functor $$F^*$$, which returns, for a set $$X$$, a fixpoint of the functor $G Y = X + F Y.$

Let’s assume that the fixpoint is to be intended as inductive, i.e. as an initial algebra. Therefore, we get, for all objects $$X$$, an initial algebra: $X + F (F^* X) → F^* X.$

Of course, those initial algebras might not exist, but they do if we choose $$F$$ carefully. For example, if $$F$$ is polynomial, then all the functors $$G$$ above are also polynomial, thus they have initial algebras.

In general, if we assume that those initial algebras all exist, then we can prove that the resulting functor $$F^*$$ is a monad, and is indeed the algebraically free monad over $$F$$.

We will first show that $$F^*$$ allows us to define a left adjoint $$L$$ for the forgetful functor $$U : \mathsf{Alg}_F → \mathsf{Set}$$. In fact, for any set $$X$$, let the carrier of $$L X$$ be $$F^* X$$, and define the algebra morphism by restriction from the initial algebra structure on $$F^* X$$: $F (F^* X) → X + F (F^* X) → F^* X.$

By definition, $$F^* X$$ is the initial object in the category of algebras of the functor $$Y ↦ X + F Y$$. Moreover, it is easy to see that the latter category is equivalent to the comma category $$(X ↓ U)$$, where the equivalence maps $$F^* X$$ to the obvious morphism $$X → U L X$$. By the characterisation of adjunctions in terms of universal arrows, it follows that $$L$$ is left adjoint to $$U$$. Clearly, $$U L = F^*$$, therefore $$F^*$$ is a monad.

To conclude the proof, we need to show that the adjunction $$L ⊣ U$$ is monadic, i.e. that the comparison functor from $$F$$-algebras to $$F^*$$-algebras is an equivalence. One way to do that is to appeal to Beck’s monadicity theorem. Verifying the hypotheses is a simple exercise.

It is also instructive to look at the comparison functor as implemented in haskell:

iter :: Functor f => (f x → x) → (Free f x → x)
iter θ (Pure x) = x
iter θ (Free t) = θ (fmap (iter θ) t)


and its inverse

uniter : Functor f => (Free f x → x) → (f x → x)
uniter ψ = ψ . liftF
where liftF = Free . fmap Pure


It is not hard to prove directly, using equational reasoning, that iter θ is a monad algebra, and that iter and uniter are inverses to each other.

## Algebraically free monads are free

The documentation for Free says:

A Monad n is a free Monad for f if every monad homomorphism from n to another monad m is equivalent to a natural transformation from f to m

which doesn’t look at all like our definition of algebraically free monad. Rather, this says that $$N$$ is defined to be the free monad over $$F$$ if the canonical natural transformation $$F → N$$ is a universal arrow from $$F$$ to the forgetful functor $$\mathsf{Mon}(\mathsf{Set}) → \mathsf{Func}(\mathsf{Set}, \mathsf{Set})$$.

If that forgetful functor had a left adjoint, then we could just say that the free monad is obtained by applying this left adjoint to any endofunctor. This is actually the case if we replace $$\mathsf{Set}$$ with a so-called algebraically complete category, such as the ones modelled by Haskell, where the left adjoint is given by the (higher order) functor Free.

In $$\mathsf{Set}$$, however, we need to stick to the more awkward definition in terms of universal arrows, as not all functors are going to admit free monads. In any case, the relationship with the previously defined notion of algebraically free monad is not immediately clear.

Fortunately, we can prove that a monad is algebraically free if and only if it is free! Proving that an algebraically free monad $$F^*$$ on $$F$$ is free amounts to proving that the following natural transformation (corresponding to liftF in the Haskell code above):

$F X \xrightarrow{F \eta} F (F^* X) \to F^* X$

is universal, which is a simple exercise.

To prove the converse, we will be using Haskell notation. Suppose given a functor f, and a monad t that is free on f. Therefore, we have a natural transformation:

l :: f x → t x


and a function that implements the universal property for t:

hoist :: Monad m => (∀ x . f x → m x) → t x → m x


Now we define a functor $$\mathsf{Set}→ \mathsf{Alg}_f$$ which is going to be the left adjoint of the forgetful functor. The carrier of this functor is given by t itself, so we only need to define the algebra morphism:

alg :: f (t x) → t x
alg u = join (l u)


To show that this functor is the sought left adjoint, we have to fix a type x and an f-algebra θ : f y → y, define functions:

φ :: (x → y) → (t x → y)
ψ :: (t x → y) → (x → y)


then prove that φ g is an f-algebra morphism for all g : x → y, and that φ and ψ are inverses to each other.

The function ψ is easy to implement:

ψ a h = h . return


Defining φ is a bit more involved. The only tool at our disposal to define functions out of t x is hoist. For that, we need a monad m, and a natural transformation f → m.

The trick is to consider the continuation monad Cont y. Using θ, we define a natural transformation

w :: f z → Cont y z
w u = Cont (\k -> θ (fmap k u))


on which we can apply the universal property of t to get φ:

φ g = (runCont id) . hoist w . fmap g


From here, the proof proceeds by straightforward equational reasoning, and is left as an exercise.

## Conclusion

We looked at two definitions of “free monad”, proved that they are equivalent, and shown the relationship with the Haskell definition of Free. In the next post, we will resume our discussion of algebraic theories “with laws” and try to approach them from the point of view of free monads and monadic functors.

]]>
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