symmetric monoidal (∞,1)-category of spectra
category with duals (list of them)
dualizable object (what they have)
ribbon category, a.k.a. tortile category
monoidal dagger-category?
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 (see def. below), as discussed at commutative algebraic theory. We explore this connection below.
As a preliminary, let $V$ be a monoidal category. We say a functor $T \colon V \to V$ is strong if there are given left and right tensorial strengths
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
($B V$ is the one-object 2-category associated with a monoidal category $V$, and $(B V)^{op}$ 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 $\tilde{V} \to \tilde{V}$.
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
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 $\lambda \colon S \to T$ 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.
A strong monad $(T \colon V \to V, m \colon T T \to T, u: 1 \to T)$ (def. ) is a commutative monad if there is an equality of natural transformations $\alpha = \beta$ where
$\alpha$ is the composite
$\beta$ is the composite
Let $(T \colon V \to V, u \colon 1 \to T, m \colon T T \to T)$ be a monoidal monad, with structural constraints on the underlying functor denoted by
Define strengths on both the left and the right by
$(m \colon T T \to T, u \colon 1 \to T)$ is a commutative monad.
In fact, the two composites
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 $\alpha_{T T}$ the constraint for the composite $T T$, then by definition $\alpha_{T T}$ is the composite given by
and so, using the properties of monoidal monads, we have a commutative diagram
which completes the proof.
See here.
See examples of commutative monads.
Last revised on January 14, 2020 at 17:03:59. See the history of this page for a list of all contributions to it.