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][1], 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 what I think is the most important different, 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 this $l$ is actually a distributive law *and* is invertible, 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.


  [1]: https://ncatlab.org/nlab/show/distributive+law