Monads on a Poset

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In this post, I will briefly describe a pleasant observation I made this morning about monads on a poset, that is, a partially ordered set. In order to get there, I will have to say what exactly are monads and posets.

Monads are curious objects. I recently came across them while preparing a talk about the first few sections of Peter May’s “Geometry of Infinite Loops Spaces.” In that paper, he defines a new structure called an operad which is essentially a way to formalize tree-like compositions of maps, $f: X^{\otimes n} \rightarrow X$, for different $n$ in a some monoidal category (if you don’t know what that is, think of $(\mathsf{Vect}, \otimes, \mathbf{0})$, the category of vector spaces, with the tensor product and the zero vector space). Anyways, in May’s paper, he describes a 1-to-1 correspondence between algebras over an operad and algebras over monads. That correspondence isn’t important here, but it is important to note that monads, for some reason, seems to creep up everywhere. I don’t think they are well known amongst applied people who use a lot of fancy algebra like myself. Personally, I hadn’t been exposed to them even after a few years of reading some heavy-categorical papers in applied algebraic topology, lattice theory etc. Monads seem, however, to be fairly commonplace in certain theoretical computer science communities especially those Haskell-ites. You know who you are.

A poset, say $P$, is set with a anti-symmetric, reflexitive, and transitive binary relation, $\preceq$. Sometimes it is easier to view a poset not as a set but as a category, $\mathsf{P}$. The objects of this category are the elements of $P$. Morphisms? Given $x, y \in \mathsf{P}$,

\[\mathrm{Hom}_{\mathsf{P}}(x,y) = \begin{cases} \ast & x \preceq y \\ \emptyset & \text{otherwise} \end{cases}\]

This is kind of nice, because if you consider a functor,

\[F: \mathsf{P} \rightarrow \mathsf{Q}\]

this is just saying that $F$ is an order-preserving or monotone maps from poset $P$ to poset $Q$. It is particularly nice, when you consider an adjunction between posets,

\[F \vdash G: \mathsf{P} \longleftrightarrow \mathsf{Q}\]

This means, that

\[\mathrm{Hom}_{\mathsf{Q}} \left(F x, y \right) \cong \mathrm{Hom}_{\mathsf{P}}\left(x, G y \right)\]

But this is just,

\[\begin{align} F(x) \preceq y && \text{iff}&&&x \preceq G(y) \end{align}\]

since every arrow in a poset category is either a singleton or empty. This is called a Galois connection in the order theory literature, but really, it is just an adjunction. From an adjunction, we obtain two natural transformations, called the unit,

\[\eta: 1_{\mathsf{P}} \Rightarrow G F\]

and counit,

\(\varepsilon: FG \Rightarrow 1_{\mathsf{Q}}\) These are constructed by essentially replacing $y$ with $F x$ or replacing $x$ with $G y$ in the Hom-set definition of an adjunction and a little bit of work.

Okay. So what is a monad? Well, we already have an example, the functor $G F: \mathsf{P} \rightarrow \mathsf{P}$. Of course, we could have replaced $\mathsf{P}$ with any category and $F \vdash G$ with any adjunction. Here is the definition: a monad is a triple $(T, \mu, \eta)$ where $T$ is an endofunctor $T: \mathsf{C} \rightarrow \mathsf{C}$ , $\mu$ is a natural transformation, $\mu: T \circ T \Rightarrow T$, and $\eta$ is a natural transformation, $\eta: 1_{\mathsf{P}} \Rightarrow T$, such that $(T, \mu, \eta)$ satisfy a number of commutative diagrams describing an associativity property of $\mu$ and an identity property of $\eta$.

Proposition: Any adjunction $F \vdash G: \mathsf{C} \rightarrow \mathsf{D}$ gives rise to a monad $(G F, \eta, \mu)$.

Proof. $G F: \mathsf{C} \rightarrow \mathsf{C}$ is an endofunctor. The unit of the adjunction, $\eta: 1_{\mathsf{P}} \Rightarrow G F$, is also the unit of the monad. Let $\varepsilon$ bet the counit of the adjunction. Then, define $\mu$ as follows,

\[\mu : G F G F \Rightarrow GF\] \[\mu: GFGF x \mapsto G \circ \varepsilon FG (F x) = GF x\]

$\blacksquare$

What, then, is a monad on a poset $\mathsf{P}$? It is an endofunctor, $T: \mathsf{P} \rightarrow \mathsf{P}$, with natural transformations, $\eta: 1_{\mathsf{P}} \Rightarrow T$ and $\mu: T \circ T \Rightarrow T$. Being a functor implies $T$ is an monotone map on $P$. The transformation $\eta$ implies that $T$ is inflationary, that is, $T(x) \succeq x$. Finally, any such transformation $\mu$ implies there is a map

\[T T x \xrightarrow{\mu} Tx\]

which means that $T^2(x) \preceq T(x)$. However, $T(x) \preceq T(x’)$ and $T(x) \succeq x$ implies that $T^2(x) \preceq x$. This implies that $T^2(x) = T(x)$, or $T^2 = T$. That is, $T$ is idempotent! Hence, we have a proposition.

Proposition: A monad $(T, \mu, \eta)$ on a poset $\mathsf{P}$ is that same as a monotone, inflationary, idempotent map, $T: P \rightarrow P$.

Together with our proposition about constructing a monad from an adunction, we have the following corollary, whose proof is now trivial.

Corollary: Let $(F, G): P \longleftrightarrow Q$ be a Galois connection between posets $P$ and $Q$. Then, the following identities hold:

  • For all $x \in P$ and $y \in Q$,
\[\begin{align} G F (x) \succeq x && \text{and} && F G (y) \preceq y \end{align}\]
  • $G F GF = GF$. Furthermore, by considering the dual poset, $P^{\mathrm{op}}$, which is just reversing the arrows and the corresponding Galois connection, $G^{\mathrm{op}} \vdash F^{\mathrm{op}}: Q^{\mathrm{op}} \leftrightarrow P^{\mathrm{op}}$, we have, by this dually, $FG FG = FG$.

These are the two most important identities for a Galois connection in order theory. Yet, if we view a Galois connection from a categorical lens, this result is completely formal. I thought this observation was worth sharing and I hope you did too. Until next time.

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