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Stream: learning: questions

Topic: Not a monad, not a comonad; all he knows is what he's not.

fosco (Aug 18 2020 at 08:19):

I am in the following situation: I have defined an adjunction

$\text{L} : {\sf Cat}(\mathcal{C}^\text{op}\times\mathcal{C}, {\sf Set}) \leftrightarrows {\sf Cat}(\mathcal{C}^\text{op}\times\mathcal{C}, {\sf Set}) : \Gamma$

via a complicated coend (for $\text{L}$) and a complicated end (for $\Gamma$). It is not important to explicitly state their definition in detail now; they just arose naturally while studying something else, and they happen to enjoy a lot of interesting properties (a list of seven, and still counting), but here's an

Half-assed definition of $\Gamma G$: there is a certain functor $Q : \mathcal{C}^\text{op}\times\mathcal{C} \to {\sf Cat}(\mathcal{C}^\text{op}\times\mathcal{C}, {\sf Set})$, and I say

$\displaystyle \Gamma G := \int_A [Q(A,A)(-,-) , G(A,A)] \qquad \text{L}G := \int^A Q(A,A)(-,-)\times G(A,A)$

where, and this is very important, "the four $A$'s are treated as two", meaning that the functor

$\mathcal{C}^\text{op} \times \mathcal{C}^\text{op} \times \mathcal{C} \times \mathcal{C} \to \sf Set$

sending $(A,B,C,D)$ to $[Q(C,A)(-,-), G(B,D)]$ has been suitably precomposed with a diagonal functor

$\Delta_{\mathcal{C}^\text{op}} \times \Delta_{\mathcal C} : \mathcal{C}^\text{op} \times \mathcal{C} \to \mathcal{C}^\text{op} \times \mathcal{C}^\text{op} \times \mathcal{C} \times \mathcal{C}$

and only then we co/ended it.

Similarly for $\text{L} G$.

Of course, being a simple category theorist, whenever I see a pair of adjoint endofunctors, I suspect one of them is a monad (and, if all ${}^\text{op}$ s are on the right side, its adjoint is a comonad).

Long story short, any given $G : \mathcal{C}^\text{op}\times\mathcal{C} \to {\sf Set}$ defines a wedge

$G \to [Q(A,A)(-,-), G(A,A)]$

that in turn defines a candidate unit $\eta : G \Rightarrow \Gamma G$. Similarly, there is a cowedge $Q(A,A)(-,-)\times G(A,A) \to G$, thus a candidate counit $\sigma : \text{L} G \Rightarrow G$.

This would suggest that I can find a multiplication for $\Gamma$, and a comultiplication for $\text{L}$, but this is not the case; instead, by the way $Q$ is defined, $\Gamma$ has a comultiplication $\Gamma \Rightarrow \Gamma\Gamma$, and $\text{L}$ a multiplication $\text{L}\text{L} \Rightarrow \text{L}$.

fosco (Aug 18 2020 at 08:38):

So, to sum up, I am in the following situation:

• I have an adjunction $\text{L} \dashv \Gamma$, between endofunctors.
• $\text{L}$ is a copointed endofunctor, and $\Gamma$ is pointed.
• $\text{L}$ has an associative multiplication, and $\Gamma$ a coassociative comultiplication.

Let's call a "Balto situation" a pair of adjoint endofunctors that satisfy all the above.

Questions:

1. Are Balto situations important/studied/whatev for some reason?
2. Or rather, shall I desist from proving that $\text{L} \dashv \Gamma$ has additional structure, because a "Balto situation" tells nothing about $\text{L},\Gamma$?

Dan Marsden (Aug 18 2020 at 09:06):

Is it the case that your copoint is given by bending the wire in a string diagram on your pointed functor using the adjunction? Similarly that your comultiplication is given also by wire bending?

fosco (Aug 18 2020 at 09:13):

I would love to know! But I don't know what you mean :-)

the fact is that $Q$ as defined above is such that for every $A$ the functor $Q(A,A) : C^\text{op}\times C \to Set$ is a promonad. The promonad multiplication gives a monoid law $Q(A,A)(-,-)\times -$ and (since homming is contravariant in the first component) a comonoid law $[Q(A,A), -]$

Matteo Capucci (he/him) (Aug 18 2020 at 09:25):

Nothing useful to contribute here, but I'm curious about the naming :sweat_smile: why Balto?

Dan Marsden (Aug 18 2020 at 09:35):

In string diagrams for a bicategory such as cat, adjunctions show up as a cup and cap that satisfy the yanking equations. You can think of the adjunction as letting you bend wires up and down. Now if you have a pointed endofunctor, this is a wire coming out of a vertex, pointing downwards (say, if this is your preferred diagram convention). Using the cup, you can bend this upwards, and you get a copoint on the other side of the adjunction. You can do the same "wire bending" for the multiplication to get a comultiplication. These bendings preserve any equations that are around as well.

Dan Marsden (Aug 18 2020 at 09:36):

For example there is a well known correspondence for adjoint (co)monads that arises this way. In your case you seem to have less structure around than full (co)monads, but it seems similar if the two sides are related via the adjunction in this way.

Dan Marsden (Aug 18 2020 at 09:40):

I'm struggling to find an example picture that's not behind a paywall. If you can access "Dragging Proofs out of Pictures" or "Equational reasoning with lollipops, forks, cups, caps, snakes, and speedometers", they have the sort of diagrams I have in mind, but not the precise calculation unfortunately.

fosco (Aug 18 2020 at 09:46):

Matteo Capucci said:

Nothing useful to contribute here, but I'm curious about the naming :sweat_smile: why Balto?

https://www.youtube.com/watch?v=z4yi5P48XZI

Matteo Capucci (he/him) (Aug 18 2020 at 09:47):

Mi inchino

fosco (Aug 18 2020 at 09:56):

Dan Marsden said:

In string diagrams for a bicategory such as cat, adjunctions show up as a cup and cap that satisfy the yanking equations. You can think of the adjunction as letting you bend wires up and down. Now if you have a pointed endofunctor, this is a wire coming out of a vertex, pointing downwards (say, if this is your preferred diagram convention). Using the cup, you can bend this upwards, and you get a copoint on the other side of the adjunction. You can do the same "wire bending" for the multiplication to get a comultiplication. These bendings preserve any equations that are around as well.

I can follow you, but I fail to see the connection with my question.. :frown:

Dan Marsden (Aug 18 2020 at 10:02):

It may be what I'm saying is so obvious you've already noted it. All I was wondering is that as you have an endofunctor with a point and a multiplication, related by an adjunction to another endofunctor with a copoint and a comultiplication, is this a weakening of the relationship for adjoint (co)monads? For this to be the case, the copoint and comultiplication would have to be the ones arising via the adjunction from the other structure. I'm trying to avoid needing to understand the specifics of your situation, as it sounds interesting, but I'll then end up getting distracted from an exam I'm trying to set :) So the situation seems close to a standard one, but slightly weaker, and I'm not aware of the weaker one being studied directly.

fosco (Aug 18 2020 at 10:30):

I probably failed to make my question clear, I apologise.

My problem is that I have a functor $\Gamma$, that is a right adjoint, is pointed, but has a co multiplication instead of a multiplication. What is an algebraic structure that is half a monoid (because it has unit) but half comonoid (because it has a comultiplication)? It's neither one or the other. And yet it is something, because $\Gamma$ doesn't fail to have structure: it just intertwines it, half monoid and half comonoid.

Dually for $\text{L}$.

I have never seen such a thing; I wonder if it's useful, already known, etc. I also wonder if it's worth studying it deeper.

fosco (Aug 18 2020 at 10:33):

The "unit" of $\Gamma$ is just a pointing: how can it relate to the comultiplication, apart yielding a pointing for each iterated power $\Gamma^n$?

I know $\theta : \Gamma \Rightarrow \Gamma\Gamma$ is coassociative. What now?

Yes: I take into account the possibility that there is a counit somewhere: but have you ever tried to define a map _from_ a limit to an object? The arrow is in the wrong direction!

Matteo Capucci (he/him) (Aug 18 2020 at 10:47):

Very trivial remark: perhaps the monad and comonad arising from $L \vdash \Gamma$ inherit some interesting structure from the Balto situation.

Dan Marsden (Aug 18 2020 at 10:47):

Ah yes, sorry I hadn't picked up on the multiplication with counit aspect, that is intriguing. I'm not sure I've seen the specific situation before. It does have lots of familiar components, but combined in apparently an unfamiliar way. I see your problem, there's not many "obvious" composites about, as things are often pointing the wrong way. Hmmm.

fosco (Aug 18 2020 at 10:54):

Matteo Capucci said:

Very trivial remark: perhaps the monad and comonad arising from $L \vdash \Gamma$ inherit some interesting structure from the Balto situation.

I hope so! Another way to go might be find a triple of adjoints $U \dashv V \dashv W$ such that (say) $\text{L} = UV \dashv WV = \Gamma$

fosco (Aug 18 2020 at 11:08):

Undoubtedly, the way in which you obtain one, is dual to the way in which you obtain the other.

fosco (Aug 18 2020 at 11:08):

whoops, erased message?

Matteo Capucci (he/him) (Aug 18 2020 at 11:09):

Yeah I realized it's not great news since you need the opposite

fosco (Aug 18 2020 at 14:54):

Plot twist:

Apparently (I must admit I interleaved a couple of "what else it could be" in the proof) if the isomorphism

$\displaystyle \int^A \mathcal{C}(C', A)\times \mathcal{C}(A,B)\times \mathcal{C}(B,A)\times \mathcal{C}(A, C) \cong \mathcal{C}(C',B)\times \mathcal{C}(B,C)$

holds (with the symmetrization condition stated above), not only $\Gamma$ is a monad, but it's idempotent !

fosco (Aug 18 2020 at 14:56):

So, new question: do you find any reason why that isomorphism above is evidently true, or evidently false?

Matteo Capucci (he/him) (Aug 18 2020 at 19:59):

It has an 'holy' look

Matteo Capucci (he/him) (Aug 18 2020 at 20:04):

Quick sheets - page 69~2.png

Matteo Capucci (he/him) (Aug 18 2020 at 20:04):

I'm not sure it makes sense though

Matteo Capucci (he/him) (Aug 18 2020 at 20:05):

Also, I guess the symmetry you imposed is completely disregarded

fosco (Aug 22 2020 at 09:52):

Update: it was a monad; even idempotent.

Now: what are the algebras?