# Contents

## Definition

###### Definition

A monoidal monad is a monad in the 2-category of monoidal categories, lax monoidal functors, and monoidal transformations.

The notion of monoidal monad is equivalent to a suitable general notion of commutative monad, as discussed at commutative algebraic theory. We explore this connection below.

## Tensorial strengths and commutative monads

As a preliminary, let $V$ be a monoidal category. We say a functor $T:V\to V$ is strong if there are given left and right tensorial strengths

${\tau }_{A,B}:A\otimes T\left(B\right)\to T\left(A\otimes B\right)$\tau_{A, B} \colon A \otimes T(B) \to T(A \otimes B)
$\phantom{\rule{thinmathspace}{0ex}}$\,
${\sigma }_{A,B}:T\left(A\right)\otimes B\to T\left(A\otimes B\right).$\sigma_{A, B} \colon T(A) \otimes B \to T(A \otimes B).

which are suitably compatible with one another. The full set of coherence conditions may be summarized by saying $T$ preserves the two-sided monoidal action of $V$ on itself, in an appropriate 2-categorical sense. More precisely: the two-sided action of $V$ on itself is a lax functor of 2-categories

$\stackrel{˜}{V}:BV×\left(BV{\right)}^{\mathrm{op}}\to \mathrm{Cat}$\tilde{V} \colon B V \times (B V)^{op} \to Cat

($BV$ is the one-object 2-category associated with a monoidal category $V$, and $\left(BV{\right)}^{\mathrm{op}}$ is the same 2-category but with 1-cell composition (= tensoring) in reverse order), and the two-sided strength means we have a structure of lax natural transformation $\stackrel{˜}{V}\to \stackrel{˜}{V}$.

We remark that in the setting where $V$ is symmetric monoidal, we will assume that the left and right strengths $\tau$ and $\sigma$ are related by the symmetry in the obvious way, by a commutative square

$\begin{array}{ccc}A\otimes T\left(B\right)& \stackrel{{\tau }_{A,B}}{\to }& T\left(A\otimes B\right)\\ {}^{c}↓& & {↓}^{T\left(c\right)}\\ T\left(B\right)\otimes A& \underset{{\sigma }_{B,A}}{\to }& T\left(B\otimes A\right)\end{array}$\array{ A \otimes T(B) & \stackrel{\tau_{A, B}}{\to} & T(A \otimes B) \\ ^\mathllap{c} \downarrow & & \downarrow^\mathrlap{T(c)} \\ T(B) \otimes A & \underset{\sigma_{B, A}}{\to} & T(B \otimes A) }

where the $c$’s are instances of the symmetry isomorphism.

There is a category of strong functors $V\to V$, where the morphisms are transformations $\lambda :S\to T$ which are compatible with the strengths in the obvious sense. Under composition, this is a strict monoidal category. Monoids in this monoidal category are called strong monads.

###### Definition

A strong monad $\left(T:V\to V,m:TT\to T,u:1\to T\right)$ is commutative if there is an equality of natural transformations $\alpha =\beta$ where

• $\alpha$ is the composite

$TA\otimes TB\stackrel{{\sigma }_{A,TB}}{\to }T\left(A\otimes TB\right)\stackrel{T\left({\tau }_{A,B}\right)}{\to }TT\left(A\otimes B\right)\stackrel{m\left(A\otimes B\right)}{\to }T\left(A\otimes B\right).$T A \otimes T B \stackrel{\sigma_{A, T B}}{\to} T(A \otimes T B) \stackrel{T(\tau_{A, B})}{\to} T T(A \otimes B) \stackrel{m(A \otimes B)}{\to} T(A \otimes B).
• $\beta$ is the composite

$TA\otimes TB\stackrel{{\tau }_{TA,B}}{\to }T\left(TA\otimes B\right)\stackrel{T\left({\sigma }_{A,B}\right)}{\to }TT\left(A\otimes B\right)\stackrel{m\left(A\otimes B\right)}{\to }T\left(A\otimes B\right).$T A \otimes T B \stackrel{\tau_{T A, B}}{\to} T(T A \otimes B) \stackrel{T(\sigma_{A, B})}{\to} T T(A \otimes B) \stackrel{m(A \otimes B)}{\to} T(A \otimes B).

Let $\left(T:V\to V,u:1\to T,m:TT\to T\right)$ be a monoidal monad, with structural constraints on the underlying functor denoted by

${\alpha }_{A,B}:T\left(A\right)\otimes T\left(B\right)\to T\left(A\otimes B\right),\phantom{\rule{2em}{0ex}}\iota =uI:I\to T\left(I\right).$\alpha_{A, B} \colon T(A) \otimes T(B) \to T(A \otimes B), \qquad \iota = u I: I \to T(I).

Define strengths on both the left and the right by

${\tau }_{A,B}=\left(A\otimes T\left(B\right)\stackrel{uA\otimes 1}{\to }T\left(A\right)\otimes T\left(B\right)\stackrel{{\alpha }_{A,B}}{\to }T\left(A\otimes B\right)\right),$\tau_{A, B} = (A \otimes T(B) \stackrel{u A \otimes 1}{\to} T(A) \otimes T(B) \stackrel{\alpha_{A, B}}{\to} T(A \otimes B)),
$\phantom{\rule{thinmathspace}{0ex}}$\,
${\sigma }_{A,B}=\left(T\left(A\right)\otimes B\stackrel{1\otimes uB}{\to }T\left(A\right)\otimes T\left(B\right)\stackrel{{\alpha }_{A,B}}{\to }T\left(A\otimes B\right)\right).$\sigma_{A, B} = (T(A) \otimes B \stackrel{1 \otimes u B}{\to} T(A) \otimes T(B) \stackrel{\alpha_{A, B}}{\to} T(A \otimes B)).
###### Proposition

$\left(m:TT\to T,u:1\to T\right)$ is a commutative monad.

###### Proof

In fact, the two composites

$TA\otimes TB\stackrel{{\sigma }_{A,TB}}{\to }T\left(A\otimes TB\right)\stackrel{T\left({\tau }_{A,B}\right)}{\to }TT\left(A\otimes B\right)\stackrel{m\left(A\otimes B\right)}{\to }T\left(A\otimes B\right)$T A \otimes T B \stackrel{\sigma_{A, T B}}{\to} T(A \otimes T B) \stackrel{T(\tau_{A, B})}{\to} T T(A \otimes B) \stackrel{m(A \otimes B)}{\to} T(A \otimes B)
$\phantom{\rule{thinmathspace}{0ex}}$\,
$TA\otimes TB\stackrel{{\tau }_{TA,B}}{\to }T\left(TA\otimes B\right)\stackrel{T\left({\sigma }_{A,B}\right)}{\to }TT\left(A\otimes B\right)\stackrel{m\left(A\otimes B\right)}{\to }T\left(A\otimes B\right)$T A \otimes T B \stackrel{\tau_{T A, B}}{\to} T(T A \otimes B) \stackrel{T(\sigma_{A, B})}{\to} T T(A \otimes B) \stackrel{m(A \otimes B)}{\to} T(A \otimes B)

are both equal to ${\alpha }_{A,B}$. We show this for the first composite; the proof is similar for the second. If ${\alpha }_{T}$ denotes the monoidal constraint for $T$ and ${\alpha }_{TT}$ the constraint for the composite $TT$, then by definition ${\alpha }_{TT}$ is the composite given by

$TTX\otimes TTY\stackrel{{\alpha }_{T}T}{\to }T\left(TX\otimes TY\right)\stackrel{T{\alpha }_{T}}{\to }TT\left(X\otimes Y\right)$T T X \otimes T T Y \stackrel{\alpha_T T}{\to} T(T X \otimes T Y) \stackrel{T\alpha_T}{\to} T T(X \otimes Y)

and so, using the properties of monoidal monads, we have a commutative diagram

$\begin{array}{ccccc}& & TTX\otimes TY& \stackrel{{\alpha }_{T}}{\to }& T\left(TX\otimes Y\right)\\ & {}^{u\otimes 1}↗& {↓}^{1\otimes Tu}& & {↓}^{T\left(1\otimes u\right)}\\ TX\otimes TY& \stackrel{u\otimes Tu}{\to }& TTX\otimes TTY& \stackrel{{\alpha }_{T}T}{\to }& T\left(TX\otimes TY\right)\\ & {}^{1}↘& {↓}^{m\otimes m}& {↘}^{{\alpha }_{TT}}& {↓}^{T{\alpha }_{T}}\\ & & TX\otimes TY& & TT\left(X\otimes Y\right)\\ & & & {}^{{\alpha }_{T}}↘& {↓}^{m}\\ & & & & T\left(X\otimes Y\right)\end{array}$\array{ & & T T X \otimes T Y & \stackrel{\alpha_T}{\to} & T(T X \otimes Y) \\ & ^\mathllap{u \otimes 1} \nearrow & \downarrow^\mathrlap{1 \otimes T u} & & \downarrow^\mathrlap{T(1 \otimes u)} \\ T X \otimes T Y & \stackrel{u \otimes T u}{\to} & T T X \otimes T T Y & \stackrel{\alpha_T T}{\to} & T(T X \otimes T Y) \\ & ^\mathllap{1} \searrow & \downarrow^\mathrlap{m \otimes m} & \searrow^\mathrlap{\alpha_{T T}} & \downarrow^\mathrlap{T\alpha_T} \\ & & T X \otimes T Y & & T T(X \otimes Y) \\ & & & ^\mathllap{\alpha_T} \searrow & \downarrow^\mathrlap{m} \\ & & & & T(X \otimes Y) }

which completes the proof.

Revised on March 5, 2012 14:00:44 by Urs Schreiber (89.204.137.53)