The n-Category Café

First a note: the word “biadjunction” is generally used to mean “weak (one-variable) 2-adjunction” (i.e. an adjunction between bicategories), so we ought to avoid using it to mean “two-variable adjunction”.

Good point. Unfortunately, I have a tendency to get confused by 2-categorical terminology.

Now to the question, which is a good one! Note that already in the (2,1)-variable case, this construction can be done on any 2 of the 3 categories involved: in addition to $(A^{J^{\mathrm{op}}}, B^J) \to C$ we can get $(A, B^J) \to C^J$ and $(A^J, B) \to C^J$. I expect that the same is true in the $(n,m)$-variable case: we can choose any two of the categories to exponentiate by $J$, and if they are both in the codomain or both in the domain then one of the exponents has to be op‘d, whereas if one is in the domain and one in the codomain then they can get the same exponent. Does that count as more natural?

Yes, this is quite compelling, but I think we can hope for more. The situation feels more symmetric, although it still seems that adjunctions $(A, B^J) \to C^J$ and $(A^J, B) \to C^J$ are closely related while the other one is odd one out. Perhaps there is some 3-fold symmetry in the fact that in each case one of the adjoints is given by a (co)end and the other two are “pointwise”. I don’t know what to make of it, but perhaps there is relation between these adjunctions in terms of duality that you described in the end.

Another similar observation is that for a (2,1)-variable adjunction $(A, B) \to C$ and any functor $I \times J \to K$ there is a Day convolution adjunction $(A^I, B^J) \to C^K$. Maybe this also has an interesting generalization to the $(m,n)$ case.

it still seems that adjunctions $(A, B^J) \to C^J$ and $(A^J, B) \to C^J$ are closely related while the other one is odd one out.

I think that’s an illusion. The point is your next sentence:

Perhaps there is some 3-fold symmetry in the fact that in each case one of the adjoints is given by a (co)end and the other two are “pointwise”.

Here’s another way to describe the same structure as a polycategory with duals or a cyclic multicategory: get rid of the idea that morphisms have a “domain” or “codomain” entirely. Instead each morphism just has a list of objects called its “entries”. There is an involution $(-)^\bullet$ on the objects, and the composition operation is $M(A_1,\dots,A_n) \times M(B_1,\dots, B_m) \to M(A_1,\dots,A_{n-1},B_2,\dots,B_m)$ if $B_1 = A_n^\bullet$ (or equivalently $A_n = B_1^\bullet$). People who work with cyclic operads (one-object cyclic multicategories) are familiar with this, whereas people who think of polycategories as semantics for linear logic are familiar with the syntactic version of it (one-sided sequents $\vdash \Gamma$).

To make a polycategory into an “entries-only polycategory”, define $M(A_1,\dots,A_n)$ to be the arrows $(A_1,\dots,A_n) \to ()$ with empty target. (Or empty source; the involution mediates the difference). Conversely, given an entries-only polycategory, define an arrow $(A_1,\dots,A_n) \to (B_1,\dots,B_m)$ to be a morphism in $M(A_1,\dots,A_n,B_1^\bullet,\dots,B_m^\bullet)$. In the case of $MVar$, this corresponds to forcing all multivariable adjunctions to be “mutual right” (or mutual left).

In the entries-only version, all the exponentiation operations are the same: $M(A_1,\dots ,A_n) \to M(A_1,\dots,A_i^J, \dots ,A_j^{J^{op}},\dots,A_n)$. The apparent difference between $(A, B^J) \to C^J$ and $(A^J, B^{J^{op}})\to C$ is just an artifact of the separation of domain and codomain.

Another similar observation is that for a (2,1)-variable adjunction $(A, B) \to C$ and any functor $I \times J \to K$ there is a Day convolution adjunction $(A^I, B^J) \to C^K$. Maybe this also has an interesting generalization to the $(m,n)$ case.

Yes, I think it does. (Of course, you still need $A,B,C$ to be complete and cocomplete.) In fact, I think your functor $I \times J \to K$ only needs to be a profunctor. And at that level of generality, this observation subsumes the previous one: to get $(A, B^J) \to C^J$ use the functor $1\times J \to J$, while to get $(A^J, B^{J^{op}})\to C$ use the profunctor $J\times J^{op} \to 1$.

I wonder whether this can be regarded as some kind of “action” of $Prof$ on $MVar$.

I wonder whether this can be regarded as some kind of “action” of $Prof$ on $MVar$.

This seems plausible from the perspective of seeing multivariable adjunctions as “representable profunctors” that you mentioned in another comment. Say we have an $(m,n)$-variable adjunction presented as a profunctor $P \colon A_1 \times \ldots \times A_m \nrightarrow B_1 \times \ldots \times B_n$ and a profunctor $U \colon J_1 \times \ldots \times J_m \nrightarrow K_1 \times \ldots \times K_n$. (BTW, does itex have a better arrow than $\nrightarrow$ for profunctors?) Then it seems to me that the “Day convolution of $P$ by $U$” is a profunctor $A_1^{J_1} \times \ldots \times A_m^{J_m} \nrightarrow B_1^{K_1} \times \ldots \times B_n^{K_n}$ given by the end

$$\int_{j_1, \ldots, j_m, k_1, \ldots, k_n} P(a_1(j_1), \ldots, a_m(j_m), b_1(k_1), \ldots, b_n(k_n))^{U(j_1, \ldots, j_m, k_1, \ldots, k_n)} \text{.}$$

I think I can see that this is an $(m,n)$-variable adjunction again and that this is associative with respect to “cartesian product” of profunctors. It should be also compatible with the polycategorical composition, but I haven’t digested that enough yet.

Yes, that’s a good perspective. It seems like we are seeing a restriction to $MVar$ of some kind of action of $Prof$ on itself.

I don’t think iTeX has a better command for profunctors, but I sometimes write the HTML entity ⇸ giving $A ⇸ B$.

Va bgure jbeqf, n (0,1)-nqwhapgvba vf na bowrpg va n pngrtbel; juvyr gurer vf rffragvnyyl bar (0,0)-nqwhapgvba.

Your first answer is correct, but your second answer is wrong!

Don’t feel bad, though; I got the (0,0)-case wrong the first time too. It’s a tricky one. If no one gets it after a while I’ll give some hints.

Congratulations, Amar and David, you both got it right: the category of (0,0)-adjunctions is $Set^{op}$. (If our multivariable adjunctions pointed in the direction of the right adjoints instead of the left adjoints, it would be $Set$.)

There are several ways one can arrive at this answer, and you found two of the ones I was thinking of. Another is that a multivariable adjunction $(A_1,\dots,A_n) \to (B_1,\dots,B_m)$ can equivalently be defined as a profunctor $A_1\times\cdots\times A_n \to B_1 \times\cdots \times B_m$ that is “representable in each variable”, and a profunctor $1\to 1$ is just a set.

It’s worth noting that one isn’t technically forced to define (0,0)-adjunctions in this way; that is, one can make other choices and still get a 2-polycategory $MVar$. Because a (0,0)-ary morphism in a polycategory can’t be composed with anything else, it’s always possible to take a given polycategory and redefine there to be exactly one (0,0)-ary morphism. One can make other choices too, but the choice isn’t totally free, since as David pointed out we have to be able to compose a (0,1)-ary morphism and a (1,0)-ary morphism to get a (0,0)-ary morphism; in particular, as long as we have any of the former two, there has to be at least one (0,0)-ary morphism.

This is the minor difference between a cyclic multicategory and a polycategory with duals. Every symmetric polycategory with duals (at least, strictly involutive duals) has an underlying cyclic symmetric multicategory, this forgetful functor $U$ has both a left adjoint $L$ and a right adjoint $R$, and the adjunctions are almost equivalences: the unit $1 \to U L$ and counit $U R \to 1$ are isomorphisms, so that $L$ and $R$ are fully faithful, and the counit $L U \to 1$ and unit $1 \to R U$ are bijective on objects and act fully-faithfully on all kinds of morphisms except the (0,0)-ary ones. The composite $R U$ redefines there to be exactly one (0,0)-ary morphism as above, while $L U$ defines the (0,0)-ary morphisms to be freely generated by the composites of (1,0)-ary and (0,1)-ary ones subject to associativity.

If a cyclic multicategory happens to have a unit $\mathbf{1}$, then there is another way to make it into a polycategory with duals by taking the (0,0)-ary morphisms to be $\hom(\mathbf{1},\mathbf{1}^\bullet)$. And Amar is exactly right that while $MVar$ itself doesn’t have a unit, its full sub-polycategory on the complete and cocomplete categories does have $Set$ as a unit — at least, a unit up to equivalence of hom-categories.

Annoyingly, this breaks the otherwise completely strict aspect of the definition: except for the (0,0)-ary morphisms, $MVar$ is a strict 2-polycategory, but I haven’t been able to think of a way to define the (0,0)-ary morphisms correctly as (something equivalent to) $Set^{op}$ and retain the strictness of associativity, at least not without doing violence to the whole rest of the construction by applying some general strictification theorem for weak 2-polycategories (which I’m only presuming to exist anyway). Fortunately, since (0,0)-ary morphisms can’t be composed with anything else, for many purposes they can be safely ignored.

Also unfortunately, even the restriction of $MVar$ to complete and cocomplete categories is not representable: it doesn’t (I believe) have binary tensor products. But its further restriction to (small) complete posets does, essentially because the adjoint functor theorem for posets has no annoying solution-set conditions. We can try to categorify that by using locally presentable categories, whose adjoint functor theorem is always true, so that we at least get a multicategory with tensor products, hence a monoidal (2-)category; but since the opposite of a locally presentable category is no longer locally presentable (unless it was a poset to start with), we lose the duals (and the cotensor products) in this way.

This all leads nicely into my next post….

I believe the intuition, from game semantics, is that $A$ and $B$ are “player” and “opponent” views, which are a priori uncorrelated, and the equivalence $(A,B) \simeq (C,C^\mathrm{op})$ is a constraint: “the player and the opponent are, in fact, playing the same game”. In the end the structures obtained are equivalent, but I suppose one perspective may give a more direct translation of game semantics ideas.

Perhaps distinguishing between “categories” and “dual pairs in $\mathrm{MVar}$” sounds better, though?

Just to point out “Categorical Proof Theory of Classical Propositional Calculus”, by Bellin, Hyland, Robinson and Urban, 2006 (pdf), which is based around *-polycategories, but as far as I understand there’s still plenty of room for debate.

Intriguing, thanks for the link. But I’ll have a hard time being convinced of any such notion until I see nondegenerate examples that occur naturally in category theory, which I might be interested in even if I didn’t know or care about proof theory. (Ordinary polycategories, linearly distributive categories, and $\ast$-autonomous categories gave me the same problem, until I discovered Sup, $MVar$, and the Chu construction.)