nLab
monoidal monad

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

$${\tau }_{A,B}:A\otimes T(B)\to T(A\otimes B)$$\tau_{A, B} \colon A \otimes T(B) \to T(A \otimes B)

$$$$\,

$${\sigma }_{A,B}:T(A)\otimes B\to T(A\otimes B).$$\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

$$\tilde{V}:BV\times (BV{)}^{\mathrm{op}}\to \mathrm{Cat}$$\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 $(BV{)}^{\mathrm{op}}$ 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 $\tilde{V}\to \tilde{V}$.

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

$$\begin{array}{ccc}A\otimes T(B)& \stackrel{{\tau }_{A,B}}{\to }& T(A\otimes B)\\ {}^c\downarrow & & {\downarrow }^{T(c)}\\ T(B)\otimes A& \underset{{\sigma }_{B,A}}{\to }& T(B\otimes A)\end{array}$$\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 $(T:V\to V,m:TT\to T,u:1\to T)$ is commutative if there is an equality of natural transformations $\alpha =\beta$ where

• $\alpha$ is the composite

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

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

From monoidal monads to commutative monads

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

$${\alpha }_{A,B}:T(A)\otimes T(B)\to T(A\otimes B),\iota =uI:I\to T(I).$$\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

$${\tau }_{A,B}=(A\otimes T(B)\stackrel{uA\otimes 1}{\to }T(A)\otimes T(B)\stackrel{{\alpha }_{A,B}}{\to }T(A\otimes B)),$$\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)),

$$$$\,

$${\sigma }_{A,B}=(T(A)\otimes B\stackrel{1\otimes uB}{\to }T(A)\otimes T(B)\stackrel{{\alpha }_{A,B}}{\to }T(A\otimes B)).$$\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

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

Proof

In fact, the two composites

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

$$$$\,

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

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

$$\begin{array}{ccccc}& & TTX\otimes TY& \stackrel{{\alpha }_T}{\to }& T(TX\otimes Y)\\ & {}^{u\otimes 1}\nearrow & {\downarrow }^{1\otimes Tu}& & {\downarrow }^{T(1\otimes u)}\\ TX\otimes TY& \stackrel{u\otimes Tu}{\to }& TTX\otimes TTY& \stackrel{{\alpha }_{T}T}{\to }& T(TX\otimes TY)\\ & {}^1\searrow & {\downarrow }^{m\otimes m}& {\searrow }^{{\alpha }_{TT}}& {\downarrow }^{T{\alpha }_T}\\ & & TX\otimes TY& & TT(X\otimes Y)\\ & & & {}^{{\alpha }_T}\searrow & {\downarrow }^m\\ & & & & T(X\otimes Y)\end{array}$$\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.