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Stream: learning: questions

Topic: adjunction puzzle

John Baez (Aug 16 2022 at 15:58):

Yesterday James Dolan asked me this puzzle, and then told me the answer:

For any bicategory $B$ there's a bicategory $\mathrm{Adj}(B)$ with the same objects as $B$, where the morphisms are adjunctions in $B$, and the 2-morphisms are some well-known maps between adjunctions.

John Baez (Aug 16 2022 at 15:59):

For example if $B = \mathbf{Cat}$, $\mathrm{Adj}(B)$ is the usual 2-category where morphisms are adjoint functors.

John Baez (Aug 16 2022 at 16:00):

Puzzle. What are morphisms in $\mathrm{Adj}(\mathrm{Adj}(B))$?

Joe Moeller (Aug 16 2022 at 16:52):

You have four functors, two adjunctions. The internal-internal unit and counit actually pan out to give an internal unit and counit giving an adjunction between the left of the left and the left of the right, which of course forces the left of the right to be the same as the right of the left. So an adjoint triple.

Tobias Schmude (Aug 16 2022 at 17:09):

Furthermore iterating this argument shows that the morphisms in $\mathrm{Adj}^n(B)$ are adjoint $(n+1)$-strings!

Oscar Cunningham (Aug 16 2022 at 19:01):

It's also worth noting that you can go one step below the level of functors. Functors are themselves adjunctions of profunctors. So adjoint $n$-tuples of functors are adjoint $(n+1)$-tuples of profunctors.

Hmmm... I wonder if there's a universal way to go down a level. Does $\mathbf{Adj}$ have an adjoint?

John Baez (Aug 16 2022 at 19:46):

I don't know if $\mathrm{Adj}$ has an adjoint, but it reminds me of something else: there's an adjunction between adjunctions and monads.

• Every adjunction $R: C \to \mathsf{Set}, L: \mathsf{Set} \to C$ gives a monad on $\mathsf{Set}$,

and

• every monad $T$ on $\mathsf{Set}$ gives an adjunction of this sort where $C$ is the Eilenberg-Moore category of $T$,

and the second construction can be seen as a right adjoint of the first, if I've got my head on straight. (I think the Kleisli construction is the left adjoint. But maybe it's the other way round!)

John Baez (Aug 16 2022 at 19:57):

There should be some very general way to say this, but you need some conditions to build an "Eilenberg-Moore object" for a monad in a bicategory, I think. I guess a bunch of this was handled in Lack and Street's papers on the formal theory of monads, but I've never carefully read those.

Mike Shulman (Aug 18 2022 at 17:50):

$\rm Adj(Adj(B)$ plays an important role in the specification of modal type theories for adjoint strings, particularly cohesion. When you specify a 2-category $\cal M$ as the "mode theory" for a type theory, then every morphism in $\cal M$ gives rise to an adjoint pair of modal operators, a "positive" one (with a left universal property) and a "negative" one (with a right universal property). Thus, if you want an adjoint triple of modal operators, you can achieve that by specifying an adjoint pair of morphisms in $\cal M$.

A more category-theoretic way of saying something similar is that if $p:C\to B$ is a bifibration (i.e. both a fibration and an opfibration), then every morphism in $B$ gives rise to an adjoint pair of functors between the fibers over its domain and codomain. Therefore, an adjoint pair of morphisms in $B$ gives rise to an adjoint triple of functors between these fibers.

Kenji Maillard (Aug 19 2022 at 14:50):

Oscar Cunningham said:

It's also worth noting that you can go one step below the level of functors. Functors are themselves adjunctions of profunctors. So adjoint $n$-tuples of functors are adjoint $(n+1)$-tuples of profunctors.

Hmmm... I wonder if there's a universal way to go down a level. Does $\mathbf{Adj}$ have an adjoint?

For a (complete enough) $2$-category $\mathcal{K}$, one can construct its $2$-category of two-sided discrete fibrations $\mathbf{DFib}(\mathcal{K})$ . Given a $1$-cell $f \in \mathcal{K}(a,b)$, I think you get an adjoint pair $(f/b) \dashv (b/f)$ in $\mathbf{DFib}(\mathcal{K})$ by considering the adequate comma objects.
In the other direction, an adjunction $u \dashv v$ of two-sided discrete fibrations ($u \in \mathbf{DFib}(\mathcal{K})(a,b), v \in \mathbf{DFib}(\mathcal{K})(b,a)$), induce representably a family of adjunctions between profunctors $\mathcal{K}(z, u) : \mathcal{K}(z, a) \leadsto \mathcal{K}(z, b)$ and $\mathcal{K}(z, v) : \mathcal{K}(z, b) \leadsto \mathcal{K}(z, a)$ naturally in $z \in \mathcal{K}$, which then correspond to a family of functors $\mathcal{K}(z, a) \to \mathcal{K}(z, b)$ naturally in $z$, hence to a $1$-cell from $a$ to $b$ in $\mathcal{K}$.
This let me think that $\mathbf{Adj}(\mathbf{DFib}(\mathcal{K}))$ might be equivalent to $\mathcal{K}$ whenever the construction makes sense, providing some kind of inverse construction to $\mathbf{Adj}$.

Oscar Cunningham (Aug 21 2022 at 10:36):

Interesting. So basically take the definition of profunctors in terms of functors and repeat it in an arbitrary $2$-category. I can't quite see what this gives when you apply it to $\mathbf{Adj}(\mathbf{Cat})$, but it's not just $\mathbf{Cat}$ because there's a functor from $0$ to $1$ but if there was an object with an adjunction to both $0$ and $1$ then we could compose a functor going the other way.

Kenji Maillard (Aug 22 2022 at 09:07):

A similar argument also shows that $\mathbf{Adj}$ cannot be the underlying functor of a monad or a comonad.
If $\mathbf{adj}$ is the walking adjunction (the $2$-category with two objects and a single adjunction between these), then $\mathbf{Adj}(\mathbf{adj}) = \to$ the walking arrow $2$-category and $\mathbf{Adj}(\to) = \mathbb{2}$ the discrete $2$-category with two objects. There is no functor from $\to$ to $\mathbb{2}$ and no functor from $\mathbf{adj}$ to $\to$, so the former shows that $\mathbf{Adj}$ cannot be equipped with a comultiplication, and the latter that it cannot be equipped with a unit.

Oscar Cunningham (Aug 22 2022 at 11:24):

There are functors between those categories but they're all constant. Is that enough?

Kenji Maillard (Aug 22 2022 at 12:52):

Right, I have been too hasty. But in both cases you have to satisfy unitality/counitality equations that prevent constant functors to be valid solutions.