A new (?) way of composing monads

Question

By composition of monads, I mean given two monads $S$ and $T$, making their composite $S T$ into a monad. Or more generally, given two monoid $X$ and $Y$ in a non-symetric monoidal category, making $X \otimes Y$ into a monoid.

This is fairly classical and generally done using a distributive law, that is map $ l : TS \to ST$ that satisfies four axioms that I'll call "compatibility to the unit of $S$, to the unit of $T$, to the multiplication of $S$ and to the multiplication of $T$."

One then make $S T$ into a monoids/monads by defining $Id \to ST$ to be $I = II \overset{\epsilon_S \epsilon_T}\to ST$ and

$$(ST)(ST) \overset{S l T}{\to} SSTT \overset{\mu_S \mu_T}{\to} ST$$.

Now I've recently encountered a situation where I have a monad structure on $ST$, that is also obtained out of a map $l:TS \to ST$, but this map is not a distributive law and the way it is used to get a monad structure on $ST$ is kind of different.

I'll describe the situation below, But essentially my question are: Has anyone has seen something like this somewhere else ? - has been studied or considered somewhere ? Also, there is some general theory for distributive law: A distributive law as above is the same as the data of a lift of $S$ to a monad $S^l$ on the category of $T$-algebras, and a $S^l$-algebra (sometime called a $l$-algebra) is the same as $ST$-algebra. Is there some general theory of this kind for the situation I'm about to described (For e.g. a way to see this sort of composition in terms of the category of algebras) ? I haven't found anything really convincing yet, but I'm not sure what to look for...

Ok, so in my situation, I still have a map $l:TS \to ST$, which satisifes:

1. $l$ is actually an isomorphism.

2. $l$ satisfies 2 of the four axioms for distributive law: compatibility to the unit of $T$ and to the multiplication of $T$. So, following the terminology of the nLab, $l$ is a distributive law of the monad $T$ on the underlying endofunctor of $S$.

3. $l$ satisfies one additional axiom, which would be a consequence of the compatibility of the multiplication of $S$ with $l$, but is a little bit weaker, which says that the following two maps $SSTS \to ST$ are equal:

$$ SSTS \overset{\mu_S T S}{\to} STS \overset{l^{-1}S}{\to} TSS \overset{T \mu_S}{\to} TS \overset{l}{\to} ST $$

$$ SSTS \overset{S l^{-1} S}{\to} STSS \overset{ST \mu_S}{\to} STS \overset{Sl}{\to} SST \overset{\mu_S T}{\to} ST $$

And now, what I think is the most important difference with the usual notion of distributive law: the way it is used to build a monad structure on $ST$ is a little different: The unit law of $ST$ is the same, but now the multiplication is given by:

$$ (ST)(ST) \overset{l^{-1} ST}\to T S S T \overset{T \mu_S T}\to TST \overset{l^{-1} T}{\to} STT \overset{S \mu_T}\to ST $$

Note that if this $l$ is actually a distributive law and is invertible (as required by my hypothesis (1)), then this definition and the usual multiplication of $ST$ can be shown to be equal, but in the absence of the compatibility of $l$ with the multiplication of $S$ this is a different definition.

I claim that this indeed makes $ST$ into a monad (hope I haven't forgotten an axiom - it works on the example I care about anyway)

Example: I'll give one simple explicit example of this. It is a toy model of the situation I care about, but it is already quite interesting. I work in the cartesian monoidal category of Sets, so "monads" are just monoids (you can promote them to actual monads by looking at the endofunctor $X \times \_$ if you prefer).

I have $S = M$ any monoid and $T = L(M)$ is the monoid of finite lists of elements of $M$, with the multiplication being given by concatenation (so it does not involve the monoid structure of $M$).

$$l : L(M) \times M \to M \times L(M)$$

is the map that sends $(a_1,\dots,a_k),m$ to $a_1,(a_2,\dots,a_k,m)$ and $\varnothing,a$ to $a,\varnothing$. One easily check that $l$ satisfies all three conditions above, but crucially it is not a distributive law because it is not at all compatible to the multiplication of $M$ nor the unit of $M$.

The multiplication it defines on $M \times L(M)$ works as follow (the arrow represents the steps in the definition above):

$$m,(a_1,\dots,a_k), m',(a'_1,\dots,a'_{k'}) \to (m,a_1,\dots,a_{k-1}), a_k,m',(a'_1,\dots,a'_{k'}) \to $$ $$ (m,a_1,\dots,a_{k-1}), a_k m',(a'_1,\dots,a'_{k'}) \to m,(a_1,\dots,a_{k-1},a_k m'),(a'_1,\dots,a'_{k'}) \to $$ $$ m,(a_1,\dots,a_{k-1},a_k m',a'_1,\dots,a'_{k'}) $$

So the combined monad can be thought of as $L^+(M)$ the monoid of non-empty list of elements of $M$, where the multiplication is performed by concatenating the lists, but multiplying the last element of the left list with the first element of the right list. ( the unit element is the list $(1)$ for $ 1$ the unit of $M$, which in the definition above corresponds to the pair $(1,\varnothing)$).

More motivation: In case someone wants to see more examples, here is the general situation I was studying when I encountered this - but it is a lot less elementary. Fix $M=(M_0,\mu_M,\epsilon_M)$ a monad on a category $C$. I assume $C$ is locally presentable and $M$ is accessible so that all the construction below make sense.

It is a known result that the initial algebra for the endofunctor $M_0$ is isomorphic to the initial object in the category of $M$-algebras $X$ endowed with a map (not a morphism) $s:X \to X$. This only apply to the initial algebra, but there seems to be a close link between these two algebraic structures.

I'm calling $M_0^*$ the free monad on the underlying endofunctor $M_0$ of $M$, I'm calling $M^s$ the monad whose algebras are the $M$-algebras $X$ endowed with a "successor" function $s: X\to X$ as above (both exists thanks to our accessibility assumptions).

Refining the result above (I'm hiding a lot of details here), one can show that, one has isomorphisms relating the underlying endofunctors of these two monads :

$$ M_0^*(X) \simeq X \coprod M^s(X) $$ $$ M^s(X) \simeq M(M_0^*(X)) $$

(and yes, combining the two, we have $M_0^*(X) \simeq X \coprod M(M_0^*(X))$ which is a consequence from the fact that the initial algebra for an endofunctor $P$ satisfies $X \simeq P(X)$)

But it is not so clear how to relate them as monads... I computed several examples and the first line doesn't seem to work well regarding the monad structure, but in all cases (and I suspect something similar works in general though I'm not completely sure yet) the monad structure of $M^s(X)$ is obtained from the composite $M M_0^*$ by the process described above. The example detailed above corresponds to the special case where $M$ is the monad a sets $M(X) = M \times X$ after some computation.

Answer 1

Ok, I think I've solved the mystery, and it is a little disappointing: The point is that there is actually a distributive law lurking in the background , constructed from the $l$ in the original question, that gives the same monad structure on the composite $ST$. (Many thanks to მამუკა ჯიბლაძე and Varkor - I've realized it by trying to follow their surgestions in the comment.)

Right, so the idea is that given a $l : TS \to ST$ as in the original post, I consider the following map $\alpha: TS \to ST $ defined as the composite:

$$ TS \overset{\epsilon_S TS }\to STS \overset{l^{-1}S}\to TSS \overset{T \mu_S}\to TS \overset{l}\to ST $$

Then I claim that this $\alpha$ is a distributive law and that the monad structure on $ST$ defined from $l$ in the original post can be obtained from $\alpha$ by using the classical formula. The proof is just a bunch of diagram computation that would be hard to reproduce here (and it is a little late so I hope I didn't make any mistakes in my computation)

For example, in the case of $M$ and $L(M)$ in the original question, $\alpha$ is the map $L(M) \times M \to M \times L(M)$ defined by $\alpha((m_1,\dots,m_k),m) = (1 , (m_1,\dots,m_k m))$

It should be noted that while $l$ was invertible, $\alpha$ no longer is. So there seem to be still something interesting going on here. For example, I have the impression that one can also define a distributive law $ ST \to TS$ using something similar that produce the monad structure on $TS$ coming from the isomorphism $ST \simeq TS$ given by $l$ and the monad structure on $ST$). But in any case that solve my problem: This is "just" a distributive law with some special additional property, but I was using the wrong function $TS \to ST$.

The case of monoids: Following the suggestion of მამუკა ჯიბლაძე in the comment, here is what happens with distributive law between ordinary monoids, which I think gives a good idea of what happens in general:

In the standard case, a distributivity law of $T$ on $S$ can be thought of as a monoid $M$ that contains $S$ and $T$ as submonoids and such that each element of $M$ is written uniquely as $st$. The distributive law $TS \to ST$ itself is the function that gives you the expression as $st$ of an element of the form $ts$. You then use it to describe the multiplication $(s_1 t_1)(s_2 t_2) = (s_1 s'_2) (t'_1 t_1)$

The case described in the original post corresponds to the situation where we have two inclusions of monoids $i,j: T \to M$, and $S$ is a submonoid of $M$, and every element of $M$ can be written uniquely both as $j(t) s$ and as $s' i(t')$. The function $l$ is now the bijection that sends $(t,s)$ to $(s',t')$ where $j(t) s = s' i(t')$.

One can also use it to compute the product a bit like a distributivity law, but one has to be a bit more careful on how we do it:

$$(s_1 i(t_1)) (s_2 i(t_2)) = j(t'_1) s_1 s_2 i (t_2) =s'' i(t''_1) i(t_2) = s'' i(t''_1 t_2) $$

which corresponds exactly to the formula for the product in the original question.

The fact that this $l$ isn't compatible to unit law of $S$ exactly means that $i \neq j$, indeed $i$ and $j$ are respectively obtained as $T \overset{\epsilon_s T}\to ST$ and $T \overset{T \epsilon_s }\to TS \overset{l}{\to} ST$ so the compatibility of $l$ with $\epsilon_S$ is exactly the condition that $i=j$. The compatibility of $l$ with multiplication fails for similar reason: if one try to swap $j(t)$ with $s_1 s_2$ one can do it in a single step, but doing in two steps doesn't make sense as $j(t)s_1 s_2 = s'_1 i(t') s_2$ and then we are stuck and using $l$ to replace $(t',s_2)$ by a $(s'_2,t'')$ won't give the correct result.

But nonetheless, in this situation any "wrong side" product $i(t) s$ can be rewritten in the correct order as $s'i(t')$ by computing the product $(1 i(t)) (s i(1))$ using the formula given above, and this is how one obtains the distributive law $\alpha$ mentioned above.