4
$\begingroup$

Suppose that $C$ is a bicategory. (I only need a monoidal category, i.e. one object bicategory, but I will stick with bicategories, since theory of monads is more commonly stated in that setting). A monad $(X, A)$ in $C$ is defined as a monoid object $(A : X \rightarrow X, p^A : AA \rightarrow A, e^A: 1_X \rightarrow A)$ in $C(X, X)$. Given two monads $(X, A)$ and $(X, B)$, a distibutive law is a 2-cell $d : BA \rightarrow AB$ satisfying few axioms. One of the bottom lines is that, a distributive law gives rise to a monad structure on the composite $AB$. We also have the following characterization:

Theorem: There is a one-to-one correspondence between distributive laws between monads $(X, A)$ and $(X, B)$ and those monad structures on $AB$ for which $e^A1_B : B \rightarrow AB$ and $1_Ae^B : A \rightarrow AB$ are monoid morphisms and $p^{AB}(1_Ae^Be^A1_B) = 1_{AB}$.

Proof: From a distrubutive law one defines $p^{AB} = (p^Ap^B)(1_Ad1_B)$ and $e^{AB} = e^Ae^B$. Vice versa, from a monad structure on $AB$ one defines $d = (e^A1_{BA}e^B)$.

Suppose now that $(X, B)$ is a monad, but $A$ is only an endomorphism $X \rightarrow X$. Then we can ask a question:

Q: What is a structure corresponding to a monad structure on $AB$ compatible in a certain sense with the monad structure on B?

The following is an answer:

A: Consider a structure consisting of 2-cells $t : ABA \rightarrow AB$ and $k : 1_X \rightarrow BA$ satisfying:

  1. $(1_Ap^B)(t1_B)(1_{AB}t) = p(Ap^BA)(tBA)$
  2. $(Ap^B)(tB)(kAB) = 1_{AB}$
  3. $t(kA) = Ae^B$.

Then, we have a one-to-one correspondence: Given $t$ and $k$ we define $p^{AB} = (1_Ap^B)(t1_B)$ and $e^{AB} = k$. Visa versa, from a (good enough) monad structure on $AB$ we define $t = p^{AB}(1_{ABA}e^B)$ and $k = e^{AB}$.

My question is: Does anyone know an interesting example of such a structure?

In particular, does something like this come up interestingly in the context of operads? (Thinking of operads as monads). In this situation, from an operad $B_n$, a collection $A_n$ and an extra structure we get a new operad. I guess if we think of $B_n$-s as objects of operations, we can think of $A_n$-s as objects of parameters which can not be composed on their own.

$\endgroup$
1
  • 3
    $\begingroup$ +1 for lots of details, definitions, theorems... $\endgroup$ Jul 24, 2015 at 0:40

1 Answer 1

1
$\begingroup$

This gets used in homotopy theory. For example, suppose your endofunctor $F$ is a path object functor (used to construct functorial factorizations). Then the distributive law in your question is sometimes called compatibility with the monad $T$ and as you observe it allows you to lift $F$ to a path object functor on the category of $T$-algebras. This is one way of solving the difficult problem of lifting a model structure to $T$-algebras. Examples include topological spaces, chain complexes, simplicial abelian groups, and other model categories with nice fibrant replacement and with functorial path objects. Berger-Moerdijk studied operads and colored operads and their conditions about interval objects are precisely to lift the path object to algebras over these (colored) operads. See Axiomatic Homotopy Theory for Operads, especially Section 4. In general, it's easiest to have the compatibility if the operad is $\Sigma$-cofibrant, which should be thought of as a condition to say that the symmetric group actions are free (from the eyes of homotopy theory). However, they also have some results that say that if the model category is nice enough then operad-algebras get a model structure for any operad. So that gives you a whole host of examples when the distributive law was used. Donald Yau has also made good use of this lifting. In his 2009 paper with Mark Johnson they do this for monads $T$ coming from a PROP (see Lemma 3.3). In the book they are currently writing it's even more general, and a version of that general one is in my paper with Yau on right Bousfield localization, which we hope to finish soon.

$\endgroup$
1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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