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Stream: theory: category theory

Topic: Monads on Monoids

Oscar Cunningham (Apr 15 2020 at 16:03):

I just started a blog! The first post is about monads on monoids (seen as single object categories). It features a surprise appearance from the following integer sequence: 1, 1, 2, 1, 3, 2, 4, 1, 4, 3, ...

Morgan Rogers (he/him) (Apr 15 2020 at 16:23):

Haha nice hook, you caught my interest. Where does your terminology "modules" as the objects of the Eilenberg-Moore category come from?

Oscar Cunningham (Apr 15 2020 at 16:31):

If you view a monad as a monoid internal to the category of endofunctors, then it acts on the modules in the same way that a ring acts on its modules. See: https://ncatlab.org/nlab/show/module+over+a+monad. I think 'algebra' is an earlier term, but people are trying to switch over to module because it makes this analogy more consistent. A module is something which is acted on, whereas an algebra is something that does some acting.

Morgan Rogers (he/him) (Apr 15 2020 at 16:39):

Cool, thanks. This seems pretty consistent.
((People need to be careful when lifting terminology from ring theory, though; it's caused me a lot of trouble recently because "flatness" wrt tensoring of modules over a ring can be described as either preserving monos or equivalently as preserving finite limits. When generalising to tensors of actions of a monoid M, semigroup theorists chose the former as their naming convention (flat = tensoring preserves monos), but this isn't the same as preserving finite limits for these objects, so it clashes with "flat = functor preserving finite limits" in CT.))

Reid Barton (Apr 15 2020 at 16:47):

If anything this is abandoning the terminology from algebra: an algebra over a ring (or Lie algebra, etc.) is an algebra/module for a certain monad, but it's also a thing which also has its own notion of a module (module over an algebra, etc.). So this could get confusing.

Oscar Cunningham (Apr 15 2020 at 16:48):

Well, both algebras and modules are modules/algebras of monads.

Reid Barton (Apr 15 2020 at 16:52):

What I mean is, as far as I know, the term "algebra" of a monad comes from algebras of operads, which are things like E_n-algebras, or Lie algebras, etc., and in that context you couldn't "backport" the term module since it would cause terrible confusion.

Robin Piedeleu (Apr 16 2020 at 11:38):

Really interesting blog post! I like that there's already a hint of splitting a monad into an adjunction with the example of conjugation. Looking forward to the next one.

Oscar Cunningham (Apr 16 2020 at 16:12):

@Robin Piedeleu Thanks!

Oscar Cunningham (Apr 16 2020 at 16:12):

Next post: Adjunctions between Monads

Oscar Cunningham (Apr 16 2020 at 16:13):

It turns out that we don't get the situation that we get with categories, where each monad can have many adjunctions that compose to give it. Each monad on a monoid corresponds to a unique adjunction between monoids!

Johannes Drever (Apr 17 2020 at 14:46):

@Oscar Cunningham quick question regarding the wikipedia article on closures. In the section on partially ordered sets it says, that extensiveness, increasingness and idempotency may be summarized as "$x \leq cl(y)$ if and only if $cl(x) \leq cl(y)$". This looks pretty close to an adjunction and since every Galois connection gives rise to a closure operator I would expect an adjoint relation. However this would rather look like "$x \leq cl(y)$ if and only if $cl(x) \leq y$". Am I missing something?

sarahzrf (Apr 17 2020 at 14:52):

not all monads are self-adjoint, that's all

sarahzrf (Apr 17 2020 at 14:52):

in fact, pretty few are

sarahzrf (Apr 17 2020 at 14:54):

if you want to get an adjunction, you can use the kleisli or eilenberg-moore categories of the monad—there are canonical ways of factoring the monad back into an adjunction through them

sarahzrf (Apr 17 2020 at 14:56):

if the monad is idempotent, then they coincide and the right adjoint is fully faithful, so you basically get a subcategory—an example of this is Ab ↪ Grp, with the monad of the abelianization adjunction

sarahzrf (Apr 17 2020 at 14:56):

and in fact every monad on a poset is idempotent, and you get something that's probably what you want :)

sarahzrf (Apr 17 2020 at 14:57):

it's quite simple, in fact—let me give you the explicit construction

sarahzrf (Apr 17 2020 at 14:58):

let P be a proset (preordered set), and say M : P → P is a monad on P (i.e., a closure operator)

sarahzrf (Apr 17 2020 at 14:59):

then define $P^M \triangleq \{x \in P \mid M(x) \le x\}$

sarahzrf (Apr 17 2020 at 15:01):

it's straightforward to see that we also have $x \in P^M \iff x \simeq M(x)$ (i'm writing $\simeq$ instead of $=$ to denote the standard equivalence relation of "mutually ≤", but it is equality if the proset is actually a poset)

sarahzrf (Apr 17 2020 at 15:02):

so for example, if P is the power set of a topological space and M is topological closure, then P^M is the closed sets

sarahzrf (Apr 17 2020 at 15:04):

now, note that $M(x) \le M(M(x))$ is one of the laws of a closure operator, so $M(x) \in P^M$ always

sarahzrf (Apr 17 2020 at 15:05):

hence we can restrict M's codomain and get $L : P \to P^M$

sarahzrf (Apr 17 2020 at 15:06):

and then we have $L \dashv \iota$, where $\iota : P^M \to P$ is just the inclusion

sarahzrf (Apr 17 2020 at 15:06):

and $\iota \circ L = M$, so this factors M back into an adjunction

Oscar Cunningham (Apr 17 2020 at 15:34):

@Johannes Drever Yep, what sarahzrf said. Every closure operation arises from a Galois connection, but that Galois connection isn't necessarily $\mathrm{cl}:P\leftrightarrows P:\mathrm{cl}$, which is what we would need to get '$x\leq\mathrm{cl}(y)$ if and only if $\mathrm{cl}(x)\leq y$'.

sarahzrf (Apr 17 2020 at 15:36):

arises from probably more than one galois connection, even :)

Johannes Drever (Apr 17 2020 at 15:39):

@sarahzrf thanks for the explanation and the construction. I think I have to study going back and forth between adjunctions and monads a bit.

sarahzrf (Apr 17 2020 at 15:39):

basically:

sarahzrf (Apr 17 2020 at 15:40):

• taking the monad of an adjunction is a lossy operation—many distinct adjunctions may give the same monad

sarahzrf (Apr 17 2020 at 15:41):

• given a particular monad M : C → C, there is a whole category whose objects are the adjunctions that factor M (so in each of these adjunctions, the left adjoint has domain C, but its codomain is probably different for different adjunctions), and whose morphisms are functors between the codomains of the left adjoints that make certain diagrams commute

sarahzrf (Apr 17 2020 at 15:43):

• you can always construct adjunctions factoring M thru the kleisli category of M and thru the eilenberg-moore category of M, and the kleisli adjunction will be the initial object of the aforementioned category, while the eilenberg-moore adjunction will be terminal

sarahzrf (Apr 17 2020 at 15:44):

• an adjunction is "monadic" if it is terminal in the category of adjunctions with the same monad—i.e., if it is isomorphic, as an adjunction, to the eilenberg-moore adjunction for its monad

sarahzrf (Apr 17 2020 at 15:44):

• going between monads and adjunctions is itself an adjunction... or so i've heard :explosion:

Johannes Drever (Apr 17 2020 at 15:49):

sarahzrf said:

• going between monads and adjunctions is itself an adjunction... or so i've heard :explosion:

Yes, it's also in the nLab like this:

Moreover, passing from adjunctions to monads and back to their monadic adjunctions constitutes itself an adjunction between adjunctions and monads, called the semantics-structure adjunction.

That's truly amazing. So going from adjunctions to monads is a forgetful functor and going from monads to adjunctions is a free construction, right? :octopus:

sarahzrf (Apr 17 2020 at 15:50):

i believe so

sarahzrf (Apr 17 2020 at 15:51):

well, actually it looks like the left adjoint is taking the monad of an adjunction :/

sarahzrf (Apr 17 2020 at 15:51):

so shrug

sarahzrf (Apr 17 2020 at 15:52):

not every adjunction easily fits super well into "free"/"forgetful"

Johannes Drever (Apr 17 2020 at 15:59):

Good to know that "free" and "forgetful" only heuristics that don't always fit.

Oscar Cunningham (Apr 17 2020 at 16:07):

My guess would be that taking the monad of an adjunction is left adjoint to the Eilenberg-Moore construction and right adjoint to the Kleisli construction

Johannes Drever (Apr 17 2020 at 16:18):

That was a nice tangent on the topic you did not want to talk about in the article :grinning:

Morgan Rogers (he/him) (Apr 17 2020 at 17:25):

Oscar Cunningham said:

My guess would be that taking the monad of an adjunction is left adjoint to the Eilenberg-Moore construction and right adjoint to the Kleisli construction

So in fact this does fit into the usual paradigm? The construction of the category of free algebras (Kleisli) is the "free" construction relative to the forgetful (adjunction $\mapsto$ monad) functor; the Eilenberg-Moore category is the "cofree" thing? :stuck_out_tongue:

Oscar Cunningham (Apr 17 2020 at 18:08):

Sounds right!

Mike Shulman (Apr 17 2020 at 19:19):

Yes, that's basically right. I think it's reasonable to say that the Kleisli adjunction is the "free adjunction generated by a monad" -- the objects of the Kleisli category of $T$ are "all and only those that have to be there in an adjunction that generates the monad $T$", namely the free objects generated by objects of the base category, and similarly the morphisms between them are all and only those that have to be there. Similarly, the EM-adjunction is the cofree adjunction generated by a monad.