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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?

Shachaf
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    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 (http://en.wikipedia.org/wiki/Closure_operator), and the closure operators on posets you get in this way include the ones coming from adjunctions (http://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 03:20
  • @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 03:42
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    @Zach: nope. At some point I might write a blog post. – Qiaochu Yuan Jul 15 '13 at 04:49
  • Note that if $M$ is a group, then a monad in $M$ is an inner automorphism of $M$, together with the conjugating element. – Alexey Jun 19 '16 at 08:08

2 Answers2

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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.

Zach L.
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I wrote some blog posts about Monads on Monoids and Adjunctions between Monoids. The main results are:

  • If $M$ is commutative, a group, or finite, then every monad on it is a conjugation.
  • Every monad on monoids arises from a unique adjunction between monoids.
  • There are examples of monads on monoids that aren't given by conjugation. These are necessarily quite complicated, since by the first bullet point they must be infinite noncommutative monoids that aren't groups. One example is given by taking $M$ to be the monoid of order preserving functions $\mathbb N \to\mathbb N$ which are eventually translations, i.e. with the property that for all sufficiently large $n\in\mathbb N$ they have the form $n\mapsto n+c$ for some $c\in\mathbb Z$. We then define $\eta$ by $\eta(n)=n+1$, $\mu$ by $\mu(n+1)=n$ and $\mu(0)=0$, and $T$ by $T(f)(n+1) = f(n)+1$ and $T(f)(0)=0$.
Oscar Cunningham
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