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Stream: theory: category theory

Topic: Convolution of monad with a comonad

Jean-Baptiste Vienney (Dec 01 2023 at 21:06):

If $(\mathcal{C},\otimes,\multimap)$ is a closed monoidal category, $(A,\Delta,\eta)$ a comonoid in $\mathcal{C}$ and $(A,\nabla,\epsilon)$ a monoid in $\mathcal{C}$, then $A \multimap B$ is a monoid in $\mathcal{C}$. The multiplication $(A \multimap B) \otimes (A \multimap B) \rightarrow A \multimap B$ is given by starting with $f \otimes g:A \otimes A \rightarrow B \otimes B$, precomposing by $\Delta$ and postcomposing by $\nabla$. From this idea, you can get the multiplication map $(A \multimap B) \otimes (A \multimap B) \rightarrow A \multimap B$ but honnestly it's much easier to desribe it using some sequent calculus than the vocabulary of closed monoidal categories.

In fact, I believe that everything works, I haven't verified it. With some appropriate string diagrams, it should be easier than with equations.

Question: Do you have a reference for string diagrams for closed monoidal (symmetric) categories?

Now, if this works, I'd like trying to apply it to monads/comonads. I guess that $Cat, \times, Nat[-,-]$ is a closed monoidal (non symmetric) category, right?

Question: If $F:\mathcal{A} \rightarrow \mathcal{B}$ and $G:\mathcal{C} \rightarrow \mathcal{D}$ are two functors, then what is $Nat[F,G]$? I think it makes sense, right?

Jean-Baptiste Vienney (Dec 01 2023 at 21:07):

If I have a comonad $S$ and a monad $T$, I then want to obtain a monad $Nat[S,T]$ with multiplication induced
$S \rightarrow SS\overset{\alpha \bullet \beta} \rightarrow TT \rightarrow T$ given two natural transformations $\alpha,\beta:S \rightarrow T$.

Well, I'm open to any comments about this.

(It's still something I'm interested about thinking to distributions in closed monoidal differential categories where I want to obtain a monad $M$ with $MA$ a monoid for every $A$ starting from a comonad $S$ on a monoidal category such that for every objects $SA$ is a comonoid and a monad $T$ such that every $TA$ is a monoid. (Co)multiplications of (Co)Monads are about composition of smooth functions and the (Co)monoids are about multiplications of smooth functions)

Mike Shulman (Dec 01 2023 at 21:10):

If by Nat you mean the set of natural transformations, then it doesn't make sense to write $\mathrm{Nat}[F,G]$ unless $F$ and $G$ have the same domain and codomain.

Jean-Baptiste Vienney (Dec 01 2023 at 21:10):

Ok thanks, but in my case they have as I'm interested by monads and comonads

Mike Shulman (Dec 01 2023 at 21:11):

Ok, so then you're just working in the non-symmetric monoidal category of endofunctors of some fixed category $C$.

Jean-Baptiste Vienney (Dec 01 2023 at 21:12):

Ok, and is it a monoidal closed category?

Mike Shulman (Dec 01 2023 at 21:12):

That's not a closed category in general: $\mathrm{Nat}[F,G]$ isn't another endofunctor. But there are some cases when at least some internal-homs do exist in it.

Jean-Baptiste Vienney (Dec 01 2023 at 21:19):

And what are these internal-homs, if they are not of the form $Nat[F,G]$?

Jean-Baptiste Vienney (Dec 01 2023 at 21:24):

Just to clarify, $Cat[\mathcal{C},\mathcal{C}]$ is a category with objects functors and morphisms natural transformations between them.

Jean-Baptiste Vienney (Dec 01 2023 at 21:25):

Now I would want a functor $Cat[F,G]:Cat[\mathcal{C},\mathcal{C}] \rightarrow Cat[\mathcal{C},\mathcal{C}]$ given two functors $F,G:\mathcal{C} \rightarrow \mathcal{C}$

John Baez (Dec 01 2023 at 21:48):

Well you've definitely got that, by precomposition with $F$ and postcomposition with $G$ - which act on functors from $\mathcal{C}$ to itself, but also on natural transformations between such functors, via [[whiskering]].

Jean-Baptiste Vienney (Dec 01 2023 at 21:50):

Ok, perfect, so now @Mike Shulman says that it doesn't define a closed monoidal category. I'd be happy to read more explanations about that.

John Baez (Dec 01 2023 at 21:50):

Note there's nothing special about $Cat$ here: for any 2-category $\mathbf{C}$ and object $\mathcal{C} \in \mathbf{C}$, given two endomorphisms $F,G: \mathcal{C} \to \mathcal{C}$ you get a functor $\mathbf{C}(\mathcal{C}, \mathcal{C}) \to \mathbf{C}(\mathcal{C}, \mathcal{C})$ given in the same way: by pre- and post-composition, and whiskering.

Jean-Baptiste Vienney (Dec 01 2023 at 21:54):

So the question is to know if there is an adjunction
$F \times - \dashv Cat[F,-]$

Jean-Baptiste Vienney (Dec 01 2023 at 22:15):

Well, it's written on the nlab that $Cat$ is monoidal closed. So there is no problem and I think that this notion of convolution of a monad with a comonad makes sense.

Notification Bot (Dec 01 2023 at 22:15):

Jean-Baptiste Vienney has marked this topic as resolved.

Jean-Baptiste Vienney (Dec 01 2023 at 22:17):

If people always try to prove me that I'm wrong rather than helping me I'll finish not to speak with people anymore and just do math alone.

Jean-Baptiste Vienney (Dec 01 2023 at 22:59):

Sorry for that

Notification Bot (Dec 01 2023 at 22:59):

Jean-Baptiste Vienney has marked this topic as unresolved.

Jean-Baptiste Vienney (Dec 01 2023 at 22:59):

$Cat$ is monoidal closed but I'm talking about another category.

Jean-Baptiste Vienney (Dec 01 2023 at 23:00):

It doesn't change that I feel that people always try to prove me that it's not interesting or only emphasize the mistakes I make each time I try to introduce a new idea. And after some time they realize that I'm not completely dumb and just not say anything anymore. Or maybe I'm just crazy.

Mike Shulman (Dec 01 2023 at 23:34):

Jean-Baptiste Vienney said:

Now I would want a functor $Cat[F,G]:Cat[\mathcal{C},\mathcal{C}] \rightarrow Cat[\mathcal{C},\mathcal{C}]$ given two functors $F,G:\mathcal{C} \rightarrow \mathcal{C}$

Why is that what you would want? The internal-hom of two objects of a closed monoidal category is another object of that category, so if $F,G \in \mathrm{Cat}[C,C]$ we should also have $F\multimap G \in \mathrm{Cat}[C,C]$.

Jean-Baptiste Vienney (Dec 01 2023 at 23:47):

Yes, you're right, sorry.

Mike Shulman (Dec 01 2023 at 23:52):

The ones of these that do sometimes exist are right Kan extensions. Because the tensor product in ${\mathrm{Cat}}[C,C]$ is composition, the universal property of a (right) internal-hom would be $Nat[H, F\multimap G] \cong Nat[H\circ F, G]$, which is the universal property of a right Kan extension of $G$ along $F$. Using the "pointwise" formula for computing such Kan extensions, one can show that they exist in certain good cases, e.g. I think it suffices if $C$ is locally presentable and the endofunctors are accessible.

Morgan Rogers (he/him) (Dec 02 2023 at 09:03):

Jean-Baptiste Vienney said:

If people always try to prove me that I'm wrong rather than helping me I'll finish not to speak with people anymore and just do math alone.

I don't really understand why you said this, but I am sorry that you feel that way.

In your original post you mentioned symmetric monoidal closed categories. Do you actually use the symmetry somewhere in constructing the "convolution"? I don't think you do, but if I'm wrong then that could be an obstruction to using this construction in a category of endofunctors even when some exponentials exist.

On the other hand, if the category you're working over has closed structure, then you can define a functor $[F,G] : \mathcal{C}^{\mathrm{op}} \times \mathcal{C} \to \mathcal{C}$ via $(X,Y) \mapsto [FX,GY]$. The contravariance indicates a potential obstacle here: what you might like to do is compose this with a diagonal map, but due to contravariance this isn't necessarily possible. A case where it can work is when $\mathcal{C}$ has an involution... is that relevant to the situation you had in mind?

Nathaniel Virgo (Dec 02 2023 at 09:10):

Jean-Baptiste Vienney said:

Question: Do you have a reference for string diagrams for closed monoidal (symmetric) categories?

Baez and Stay's Physics, Topology, Logic and Computation: A Rosetta Stone has a string diagram calculus for closed monoidal categories.

Ralph Sarkis (Dec 02 2023 at 09:15):

Also in Chapter 3 of this tutorial.

Jean-Baptiste Vienney (Dec 02 2023 at 14:38):

Morgan Rogers (he/him) said:

Jean-Baptiste Vienney said:

If people always try to prove me that I'm wrong rather than helping me I'll finish not to speak with people anymore and just do math alone.

I don't really understand why you said this, but I am sorry that you feel that way.

In your original post you mentioned symmetric monoidal closed categories. Do you actually use the symmetry somewhere in constructing the "convolution"? I don't think you do, but if I'm wrong then that could be an obstruction to using this construction in a category of endofunctors even when some exponentials exist.

On the other hand, if the category you're working over has closed structure, then you can define a functor $[F,G] : \mathcal{C}^{\mathrm{op}} \times \mathcal{C} \to \mathcal{C}$ via $(X,Y) \mapsto [FX,GY]$. The contravariance indicates a potential obstacle here: what you might like to do is compose this with a diagonal map, but due to contravariance this isn't necessarily possible. A case where it can work is when $\mathcal{C}$ has an involution... is that relevant to the situation you had in mind?

I haven't been very clear. I think, this is true:
Screenshot-2023-12-02-at-9.33.50AM.png

As a generalization of this:
Screenshot-2023-12-02-at-9.35.19AM.png

From the nlab page convlution algebra. I'm the person who wrote the first screenshot, and at the time it seemed obvious to me that it should work but I must verify it with string diagrams. So, thank you Nathaniel and Ralph for the references.

Jean-Baptiste Vienney (Dec 02 2023 at 14:41):

If Proposition 1.1 is true, then I would want to apply to the category of endofunctors $End(\mathcal{C})$ of an arbitrary category $\mathcal{C}$, with tensor product the composition, given a comonad $A$ and a monad $B$. But as Mike Shulman said, this is not a monoidal closed category in general, so we can't apply it directly.

Jean-Baptiste Vienney (Dec 02 2023 at 14:44):

Morgan Rogers (he/him) said:

Jean-Baptiste Vienney said:

If people always try to prove me that I'm wrong rather than helping me I'll finish not to speak with people anymore and just do math alone.

I don't really understand why you said this, but I am sorry that you feel that way.

In your original post you mentioned symmetric monoidal closed categories. Do you actually use the symmetry somewhere in constructing the "convolution"? I don't think you do, but if I'm wrong then that could be an obstruction to using this construction in a category of endofunctors even when some exponentials exist.

On the other hand, if the category you're working over has closed structure, then you can define a functor $[F,G] : \mathcal{C}^{\mathrm{op}} \times \mathcal{C} \to \mathcal{C}$ via $(X,Y) \mapsto [FX,GY]$. The contravariance indicates a potential obstacle here: what you might like to do is compose this with a diagonal map, but due to contravariance this isn't necessarily possible. A case where it can work is when $\mathcal{C}$ has an involution... is that relevant to the situation you had in mind?

No I don't use the symmetry. I mentioned symmetric monoidal closed categories because I'm confused about non-symmetric monoidal closed categories which can be left-closed, right-closed or biclosed.

Todd Trimble (Dec 02 2023 at 14:57):

So regarding the first screenshot, the thing that comes to mind is to define a composition

$(A \multimap B) \otimes (A \multimap B) \to (A \otimes A) \multimap (B \otimes B) \overset{\delta \multimap \mu}{\to} A \multimap B$

where $\delta$ denotes comultiplication and $\mu$ denotes multiplication. The only obstacle is to define the first map, which seems to require a symmetry. This is similar to what happens in enriched category theory, where you can tensor two $V$-categories if you have a symmetry or a braiding, but otherwise not.

Jean-Baptiste Vienney (Dec 02 2023 at 15:33):

Yes, you're right, I think you indeed need a symmetry. Thanks.

Todd Trimble (Dec 02 2023 at 15:54):

All that being said, your original comment seems okay, that we have a composition

$\mathrm{Nat}(S, T) \times \mathrm{Nat}(S, T) \to \mathrm{Nat}(SS, TT) \overset{\mathrm{Nat}(\delta, \mu)}{\longrightarrow} \mathrm{Nat}(S, T)$

where the first map is a Godement interchange. So the set $\mathrm{Nat}(S, T)$ becomes an ordinary monoid, if $S$ is a comonad and $T$ is a monad (both on the same category, of course). I don't think I've ever thought about that before, but that might useful for some occasions.

Nathanael Arkor (Dec 02 2023 at 16:53):

This structure looks like one of the operations arising in the Sweedler theory of monads: see the "Sweedler copower of a monoid" on page 7 of McDermott–Rivas–Uustalu's Sweedler Theory of Monads.

Jean-Baptiste Vienney (Dec 02 2023 at 17:04):

Thank you, this paper looks super interesting!