Category Theory
Zulip Server
Archive

You're reading the public-facing archive of the Category Theory Zulip server.
To join the server you need an invite. Anybody can get an invite by contacting Matteo Capucci at name dot surname at gmail dot com.
For all things related to this archive refer to the same person.

Stream: learning: questions

Topic: Profunctor and Categorified Linear Algebra

Peva Blanchard (Mar 10 2024 at 10:03):

I am still "meditating" on this article on n-category café by @Simon Willerton

The article motivates the definition of the nucleus of a profunctor by building up from basic linear algebra.

Linear algebra

We have two (finite) sets $A$ and $B$, and a matrix $M : A \times B \rightarrow \mathbb{R}$ (a set function). By curryfication, and definition of free vector spaces, we get two functions (linear maps)

$$M^* : \mathbb{R}^A \rightarrow \mathbb{R}^B \\ M_* : \mathbb{R}^B \rightarrow \mathbb{R}^A$$

and they are adjoint

$$\langle M^* p , q \rangle = \langle p, M_* q \rangle = \sum_{a,b} M(a,b)p(a)q(b)$$

Then, we can consider the eigenvectors with eigenvalue $1$ of $\Delta = M^*M_*$

$$Fix(\Delta) = \{ q | \Delta q = q \}$$

Profunctor

The analogy consists in having $A$ and $B$ be $V$-enriched category, and $M : A^{op} \times B \rightarrow V$ a profunctor. By a very similar construction, we get functors

$$M^* : V^{A^{op}} \rightarrow (V^B)^{op} \\ M_* : (V^B)^{op} \rightarrow V^{A^{op}}$$

and they are adjoint

$$(V^B)^{op}(M^* p, q ) \simeq V^{A^{op}}(p, M_* q) \simeq \int_{a,b} [p(a) \otimes q(b), M(a,b)]$$

The nucleus of the profunctor can be described as the subcategory spanned by the "fixed points" of the associated monad $\Delta = M^*M_*$

$$Fix(\Delta) = \{ q | \Delta q \simeq q \}$$

Question

The direct analog of the nucleus of the profunctor in the linear algebra world would be the space of eigenvectors of $\Delta$ with eigenvalue $1$.

However, I feel left hungry for more, because we may think as the space $E_{\lambda}$ of eigenvectors with eigenvalue $\lambda$ as being a fixed point of the action of $\Delta$ on subspaces.

But I have trouble making the analogy exact: can we have all the eigenspaces of $\Delta$ being the nucleus of a well-chosen profunctor?

Peva Blanchard (Mar 10 2024 at 10:27):

ps: the analogy is quite fascinating, because in graph theory, when $M$ is the incidence matrix, then $\Delta$ is the graph laplacian

John Baez (Mar 10 2024 at 16:55):

It sounds like you're wanting to the take the analogy between profunctors and linear maps and turn it into a functor, following the principle "every analogy is yearning to become a functor". So we could hope for a functor that sends (perhaps enriched) profunctors to linear maps, or a functor that sends linear maps to (perhaps enriched) profunctors.

John Baez (Mar 10 2024 at 17:02):

I mainly know a functor that sends spans of finite groupoids to linear maps between finite-dimensional vector spaces - it's called degroupoidification. But @Todd Trimble has recently described a functor from spans of groupoids to profunctors between groupoids. (I'm not claiming it's new, just that he recently wrote about it.) So while none of this proves it, there might be an interesting way to convert profunctors between finite groupoids (or more general categories) into linear maps between vector spaces.

Peva Blanchard (Mar 10 2024 at 18:36):

Great, thank you for the references!

Peva Blanchard (Mar 10 2024 at 23:17):

Ok, this is impressive, I was not expecting such a fascinating relation between spans of groupoids, combinatorics and quantum mechanics. This forum is really amazing!

Peva Blanchard (Mar 17 2024 at 11:11):

I've continued on the "functor chasing" path. I realized that I was not really after a functor from profunctor to linear maps, but more after a functor between "ways to build an adjunction pair". E.g., on the linear algebra side, from any matrix I get two adjoint linear maps; and the profunctor side, from the $V$ hom profunctor, I get the Isbell adjunction.

So I tried to put together the pattern into, let's call it, an "Isbell situation". That would be a category $C$ with

• a special object $V$
• a monoidal product $\otimes$, and the corresponding internal hom $[\_,\_]$
• an involutive endofunctor $\star : C \rightarrow C$
• for each object $X$, a pairing $\langle\_,\_\rangle : X^{\star} \otimes X \rightarrow X \rightarrow V$
• For every $X$, $Y$:
  • A "free-completion operator" $\mu : [ X, [Y^{\star}, V]] \rightarrow [[X, V]^{\star}, [Y^{\star}, V]]$
  • A "free-co-completion operator" $\sigma: [Y, [X, V]^{\star}] \rightarrow [[Y^{\star}, V], [X, V]^{\star}]$

This data should have satisfy just enough laws so that I can replay the construction of the Isbell adjunction. I.e., the following proposition should hold.

For any "matrix data" $R : Y^{\star} \otimes X \rightarrow V$, we obtain two arrows

$$R^{\star} : [Y^{\star}, V] \rightarrow [X, V]^{\star} \\ R_{\star} : [X, V]^{\star} \rightarrow [Y^{\star}, V]$$

that are adjoints, i.e., for every $\phi : A \rightarrow [Y^{\star}, V]$ and $\psi: A \rightarrow [X,V]^{\star}$,

$$\langle R^{\star}\phi, \psi \rangle = \langle \phi, R_{\star} \psi \rangle$$

as arrows $A \rightarrow V$.

Peva Blanchard (Mar 17 2024 at 11:23):

For instance, the Isbell adjunction occurs with $C$ being the category of $V$-enriched categories

• cartesian product, and functor category as internal hom
• $V$ is the enriching category (e.g., Set)
• $\star$ is the "opposite" functor
• the pairing is $hom$, the identity profunctor
• and $\mu$ and $\sigma$ are the actual free completion and free co-completion

The linear algebra case occurs with $C$ as follows

• each object is a finite dimensional vector space $X$ with a inner product with values in $\mathbb{R}$
• product is the tensor product of vector spaces, and internal hom is the space of linear maps between two vector spaces
• special object is $\mathbb{R}$
• $\star$ is the identity
• $\mu$ and $\sigma$ are defined by "linear extension"

Peva Blanchard (Mar 17 2024 at 11:24):

Then I can consider, I guess, the category of "Isbell situations" and see how different instances relate to each other.

Peva Blanchard (Mar 17 2024 at 11:26):

Quite funnily, in the linear algebra case, the two operators $\mu$ and $\sigma$ are really the same, which is quite special.

Peva Blanchard (Mar 17 2024 at 11:55):

For instance, let's consider the Isbell situation associated with $Bool$-Cat (boolean enriched categories, i.e., preorders). If I take $X$ and $Y$ two discrete preorders, then $[X, Bool]^{\star}$ is the set of subsets of $X$ ordered by containment, and $[Y^{\star}, Bool] = [Y, Bool]$ is the set of subsets of $Y$ ordered by inclusion.
Then, $\mu$ and $\sigma$ are clearly not the same:

$$\mu (\phi_x)_{x \in X} = (A \subseteq X) \mapsto \bigcap_{x \in A} \phi_x \\ \sigma(\psi_y)_{y \in Y} = (B \subseteq Y) \mapsto \bigcup_{y \in B} \psi_y$$

Peva Blanchard (Mar 25 2024 at 16:43):

I understood that what I am looking for is not really a functor from profunctors to linear maps, but a "functor" from "ways of doing things in the profunctor world" to "ways of doing things in the linear algebra world" (or vice-versa).

I tried to factor out the common pattern (I actually deleted my previous response because of a big mistake of mine). After much wandering, I stumbled upon the notion of equipment, which I think is the setting I was looking for. There is the equipment of $V$-profunctors, and the equipment (I think) of finite dimensional linear relations. And my question becomes: Is there a "functor" between these two equipments (or variants of them)? Can the nucleus of a proarrow be defined in an equipment, and how this "functor" acts on nuclei?

I should put double quotes everywhere in the previous paragraph, because equipment, double categories and related stuff make a lot of things to absorb for me, so I'm not 100% confident in my reading. But I find the quest very exciting.

ps: there is actually a short thread of comments under the mentioned article, by @John Baez and @Mike Shulman, discussing the analogy. It was 15 years ago! so I guess some things have been sorted out by now.