Can every monad give rise to a monad transformer?

361 Views Asked by At

Can every monad give rise to a monad transformer?

In the paper Calculating monad transformers with category theory by Oleksandr Manzyuk, one finds a construction of monad transformers as translating monads along adjunctions. In particular, considering the Eilenberg-Moore construction, we can translate the monads on the category $C^T$ of $T$-algebras to monads on the original category $C,$ where $T$ is the monad we start from. Hence, if somehow a monad on $C$ induces a monad on $C^T,$ then we can translate that induced monad along the adjunction between $C$ and $C^T$ that defines $T.$ Since we can regard $C^T$ as a subcategory of $C,$ we know that there is indeed an induction if every monad $M$ on $C$ maps $T$-algebras to $T$-algebras, e.g. if $T$ commutes with all monads; in particular, the Writer-monad-transformers can be derived this way. But I know almost nothing about the center of the category $\text{Mon}(C)$ of monads on $C,$ which is equivalent to the category of monoids in the functor category $C^C=\text{Hom}(C, C).$

Then I found this question on MSE, from which I see that the functor $T\rightarrow\text{Ran}_{F^T}(F^TT)$ is a lax monoidal functor, hence inducing a functor $\text{Mon}(C)\rightarrow\text{Mon}(C^T),$ where $F^T$ is the left adjoint functor to the forgetful functor $U^T:C^T\rightarrow C.$ So the question becomes

When will the right Kan extension $\text{Ran}_{F^T}(F^TT)$ exist?
Does the construction ($T\rightarrow \text{Ran}_{F^T}(F^TT)\rightarrow\text{translating along }F^T$) give the right notion of monad transformers?

This question sounds quite obscure and confusing, but all I want to know is that are there general constructions of monad transformers for any given monad?

Any answer or reference will be sincerely appreciated, and thanks in advance.