Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've been trying to figure out what having a monad in a monoid (i.e. a category with one object) would mean.

As far as I can tell it would be a homomorphism (functor) $T : M → M$, with two elements (natural transformation components) $\eta, \mu : M$, such that

  • $\forall x. \eta x = T(x) \eta$
  • $\forall x. \mu T(T(x)) = T(x) \mu$
  • $\mu \eta$ = $\mu T(\eta)$ = 1
  • $\mu T(\mu) = \mu \mu$

The identity monad $T(x) = x$, with $\eta = \mu = 1$, is an obvious example for any monoid. But no other examples really come to mind... These laws seem a bit strange. Are there any interesting examples, or any good intuition for what the laws would mean?

share|improve this question
5  
If your goal is to get a feel for what monads are by looking at a special case, look at posets, not monoids. A monad on a poset is a closure operator (en.wikipedia.org/wiki/Closure_operator), and the closure operators on posets you get in this way include the ones coming from adjunctions (en.wikipedia.org/wiki/Galois_connection). Their closed subsets are often important and the subject of major theorems, e.g. the Nullstellensatz, the fundamental theorem of Galois theory, and Godel's completeness theorem. –  Qiaochu Yuan Jul 15 '13 at 3:20
    
@QiaochuYuan Do you have any good papers or books on the connections of those topics with monads? I would enjoy reading them! –  Zach L. Jul 15 '13 at 3:42
    
@Zach: nope. At some point I might write a blog post. –  Qiaochu Yuan Jul 15 '13 at 4:49

1 Answer 1

up vote 4 down vote accepted

So you can interpret a functor $T$ as a homomorphism $T:M \rightarrow M$. What is the interpretation of a natural transformation between two such functors? Well, it should assign to each element in the base category (there's only one!) some morphism such that the necessary diagram is natural. So we can think of $\eta, \mu$ as being elements of the monoid $M$, satisfying the following identities: $$\begin{align*}\eta m &= T(m)\eta \\ \mu T^2(m) &= T(m) \mu \end{align*}$$ for all $m$.

The associativity of $T$ becomes $$ \mu T(\mu) = \mu^2 $$ Note two things. First, that this is an identity of elements of the monoid $M$. The LHS corresponds to the natural transformation $\mu \circ T\mu$, while the RHS is the natural transformation $\mu \circ \mu T$. Since the functor $T$ fixes the only point in the category, $\mu T = \mu$. The unit law is:

$$\mu T(\eta) = e = \mu \eta.$$

It's not clear to me right now what this means for a general monoid, but for a group, we see that $T$ is actually just conjugation by $\eta$, and $\mu$ is the inverse of $\eta$. So at least for groups, the monads are just inner automorphisms, which is nice. I can't think of an interpretation for a general monoid, but the intuition from groups might help.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.