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Stream: learning: questions

Topic: Like monad is to category, so X is to operad.

Ignat Insarov (Jun 29 2025 at 21:06):

I am not sure if what I want makes sense, but what I want is a functor $T: \mathcal C^J → \mathcal C$ that is a monad in every $j ∈ J$, with some magical compatibility conditions on top. In the case of $J = \{⋆\}$, this would simply be a monad. Can I generalize monads this way?

starting point 1: equations of monad

Here are the usual equations for the identity and composition of a monad $⟨T; η; μ⟩$:

$$\begin {gather} μ ∘ ηT = \operatorname {id} = μ ∘ Tη: T → T \\ μ ∘ μ = μ ∘ Tμ \end {gather}$$

starting point 2: identity and composition in operads

In an operad, we have arrows that look like $X^J → X$, and the definitions of identity and composition are suitably altered: composition takes one function «after» and $J$ functions *«before
»*. This is technical but straightforward and intuitive. The equations for the identity and composition in some operad $\mathcal O$ look like so:

$$\begin {align} \left . ∀ n ∈ ℕ, f ∈ \mathcal O_n \quad \right | \quad & \operatorname {id} ∘ \prod_i^1 f = f = f ∘ \prod_i^n \operatorname {id} \\ \left .{\begin {aligned} ∀ n ∈ ℕ, ∀ m ∈ ℕ^n \\ f ∈ \mathcal O_n, g ∈ \prod_i^n \mathcal O_{m (i)} \\ h ∈ \prod_i ^n \prod_j^{m(i)} \mathcal O_j \end {aligned}} \quad \right | \quad & \left (f ∘ \prod_i^n f_i \right ) ∘ \prod_i^n \prod_j^{m (i)} h_{ij} = f ∘ \prod_i^n \left (g_i ∘ \prod_j^{m (i)} h_{ij} \right ) \end {align}$$

Here, the circle $∘$ is a binary operator, and the $\operatorname {id}$ is a nullary operator.

synthesis: equations of monad in operad

I am now going to transplant the identity equation for operads to the setting of functor category. Let us denote the circle as $μ$ and $\operatorname {id}$ as $η$. Let us also denote the product big operator as $T$ with suitable indexing attached. So, for example, $\mathop T _i^2 η$ would mean the product $\operatorname {id} × \operatorname {id}$. Then, we can write the same equations like this:

$$\begin {gather} μ ∘ η \mathop T = \operatorname {id} = μ ∘ \mathop T \limits_i^n η \\ μ ∘ μ = μ ∘ \mathop T \limits_i^n μ \end {gather}$$

Looks very similar to the usual equations for the identity and composition of monad! Although I see that this is a stretch. I want to believe!

example: the product co-monad

The example I have in mind is the product co-monad $⟨X; Y⟩ ↦ X × Y$ in sets, which is a monad in the opposite category. I am going to denote the arrows in the opposite category as reversed arrows $←$ and composition as circle with reversed arrow on top $\overset ← ∘$. Now, product is a monad (in the opposite category) on either side already:

$$\begin{gather} η_{\rm left}: X ← X × Y = ⟨x; y⟩ ↦ x \\ μ_{\rm left}: X × Y × Y ← X × Y = ⟨x; y⟩ ↦ ⟨⟨x; y⟩; y⟩ \end{gather}$$

$$\begin{gather} η_{\rm right}: Y ← X × Y = ⟨x; y⟩ ↦ y \\ μ_{\rm right}: X × X × Y ← X × Y = ⟨x; y⟩ ↦ ⟨x; ⟨x; y⟩⟩ \end{gather}$$

However, we do not have a single unambiguous $μ$ operation:

$$μ_{\rm left} \overset ← ∘ μ_{\rm right} ≠ μ_{\rm right} \overset ← ∘ μ_{\rm left}$$

These compositions are not even unambiguous. For example, there are two ways to have $μ_{\rm left} \overset ← ∘ μ_{\rm right}$:

$$\begin {align} μ_{\rm right} \overset ← ∘ μ_{{\rm left}, ⟨X × X, Y⟩}&: X × X × Y × Y ← X × X × Y ← X × Y \\ μ_{\rm right} \overset ← ∘ μ_{{\rm left}, ⟨X, X × Y⟩}&: X × X × Y × X × Y ← X × X × Y ← X × Y \end {align}$$

What I imagine is that we should have a combined $μ$ defined like so:

$$μ: X × Y × X × Y ← X × Y = ⟨x; y⟩ ↦ ⟨⟨x; y⟩; ⟨x; y⟩⟩$$

Now, we have the identity equations to verify:

$$\begin {alignat*} {0} μ \overset ← ∘ η_{\rm left} \mathop T &= (⟨x; y⟩ ↦ ⟨⟨x; y⟩; ⟨x; y⟩⟩) \overset ← ∘ (⟨x; y⟩ ↦ x) &= \operatorname {id} \\ μ \overset ← ∘ η_{\rm right} \mathop T &= (⟨x; y⟩ ↦ ⟨⟨x; y⟩; ⟨x; y⟩⟩) \overset ← ∘ (⟨x; y⟩ ↦ y) &= \operatorname {id} \\ μ \overset ← ∘ \mathop T \limits_i^{\{{\rm left}; {\rm right}\}} η_i &= (⟨x; y⟩ ↦ ⟨⟨x; y⟩; ⟨x; y⟩⟩) \overset ← ∘ ((⟨x; y⟩ ↦ x) × (⟨x; y⟩ ↦ y)) &= \operatorname {id} \end {alignat*}$$

The equation for composition looks like so:

$$μ \overset ← ∘ μ = μ \overset ← ∘ \mathop T \limits_i^{\{{\rm left}; {\rm right}\}} μ$$

And it holds in our example:

$$\begin {alignat*} {0} (⟨x; y⟩ ↦ ⟨⟨x; y⟩; ⟨x; y⟩⟩) \overset ← ∘ (⟨x; y⟩ ↦ ⟨⟨x; y⟩; ⟨x; y⟩⟩) \\ = ⟨x; y⟩ ↦ ⟨x; y; x; y; x; y; x; y⟩ \\ = \\ (⟨x; y⟩ ↦ ⟨⟨x; y⟩; ⟨x; y⟩⟩) \overset ← ∘ (⟨x; y⟩ ↦ ⟨⟨x; y⟩; ⟨x; y⟩⟩) × (⟨x; y⟩ ↦ ⟨⟨x; y⟩; ⟨x; y⟩⟩) \\ = ⟨x; y⟩ ↦ ⟨x; y; x; y; x; y; x; y⟩ \end {alignat*}$$

So, it kind of seems that we have an example of what I wanted here.

my question

Is this thing a thing? Does it make sense?

Ignat Insarov (Jul 01 2025 at 08:06):

Makes no sense at all?

John Baez (Jul 01 2025 at 08:18):

Here's a vague thought:

If $\mathcal{C} \mapsto \mathcal{C}^J$ is a pseudomonad on Cat, or can be made so when $J$ has enough structure, then the microcosm principle suggests that we might be able to define a kind of generalized monad $T: \mathcal{C}^J \to \mathcal{C}$ "riding" this pseudomonad.

I could be mixed up, but this is how I would start.

Ignat Insarov (Jul 01 2025 at 08:29):

What is the standard introduction to pseudomonads? Is it the same as 2-monads?

fosco (Jul 01 2025 at 08:33):

John Baez said:

Here's a vague thought:

If $\mathcal{C} \mapsto \mathcal{C}^J$ is a pseudomonad on Cat, or can be made so when $J$ has enough structure, then the microcosm principle suggests that we might be able to define a kind of generalized monad $T: \mathcal{C}^J \to \mathcal{C}$ "riding" this pseudomonad.

I could be mixed up, but this is how I would start.

"Raising to a comonoid object" is always a monad, and every $J$ in a Cartesian category is a comonoid (even in a canonical way); so $[J,-]$ is a monad. $J$ can be a discrete category, or even a nondiscrete one

fosco (Jul 01 2025 at 08:35):

if $J$ is a monoid, on the other hand, raising to $J$ is a comonad, and $T : C^J \to C$ is an arrow in its coKleisli category.

Amar Hadzihasanovic (Jul 01 2025 at 08:35):

If $(J, -\otimes-, i)$ is a strict monoidal category, it would seem that you could define some kind of monad-like structure on a functor $T\colon \mathcal{C}^J \to \mathcal{C}$ with multiplication of type

$$\mu\colon T \circ (T^J) \circ \mathcal{C}^{-\otimes-} \Rightarrow T$$

and unit of type

$$\eta\colon \mathcal{C}^i \Rightarrow T$$

which reduces to a usual monad when $J$ is the terminal category with its unique monoidal structure.

Amar Hadzihasanovic (Jul 01 2025 at 08:36):

Not sure what this is.

fosco (Jul 01 2025 at 08:36):

So in the case of $J$ monoid, the correct axioms for your T to be a monad-like gadget are precisely the monad axioms for the coKleisli endo-morphism $T^\sharp : C \rightsquigarrow C$ that T defines.

Amar Hadzihasanovic (Jul 01 2025 at 08:36):

...Probably exactly what Fosco is saying :)

fosco (Jul 01 2025 at 08:37):

Yeah, I expect unwinding my definition one gets Amar's maps

fosco (Jul 01 2025 at 08:38):

(probably my more intrinsic description makes it easier to pass from strict monoids to pseudomonoids, probably not? Anyway, it's a potato-pseudopotayto situation)

Ignat Insarov (Jul 01 2025 at 08:45):

Is there any possibility for a more accessible explanation? I have no idea what «raising to a comonoid object» means, and from then on it is an uphill battle for me.

fosco (Jul 01 2025 at 08:45):

It's simply the functor $C\mapsto C^J$

fosco (Jul 01 2025 at 08:45):

IIUC for you $J$ is a set?

fosco (Jul 01 2025 at 08:46):

so $C^J = \prod_{j\in J} C$ is a product of $|J|$ copies of C; legitimately called "C to the J"

Ignat Insarov (Jul 01 2025 at 08:47):

So, your $C$ is an object of some ambient category? And $C ↦ C^J$ is the action on objects of an exponential functor?

Amar Hadzihasanovic (Jul 01 2025 at 08:48):

We are taking specifically the exponential functor in $\mathbf{Cat}$ seen as a cartesian closed category.

fosco (Jul 01 2025 at 08:48):

by extension, more generally, if $J$ is a category, $C^J$ is the category of functors $J\to C$; this more general description matches the case when $J$ is discrete

fosco (Jul 01 2025 at 08:48):

Ignat Insarov said:

So, your $C$ is an object of some ambient category? And $C ↦ C^J$ is the action on objects of an exponential functor?

$C$ is a category; the functor $(-)^J$ is an endofunctor of the category of categories.

Ignat Insarov (Jul 01 2025 at 08:48):

So $C$ is a category seen as an object in a category of categories, alright!

Ignat Insarov (Jul 01 2025 at 08:54):

So we want $J$ to be a monoidal category, right? But discrete categories are not usually monoidal, right? So this construction cannot be done when $J$ is a non-trivial set.

Ignat Insarov (Jul 01 2025 at 08:55):

So in particular this would not explain my product example.

Ignat Insarov (Jul 01 2025 at 08:55):

Right?

John Baez (Jul 01 2025 at 08:56):

Ignat Insarov said:

So, your $C$ is an object of some ambient category? 

His $C$ is your $\mathcal{C}$. Not all of us us can afford fancy fonts. :wink:

Amar Hadzihasanovic (Jul 01 2025 at 08:57):

Ignat Insarov said:

So we want $J$ to be a monoidal category, right? But discrete categories are not usually monoidal, right? So this construction cannot be done when $J$ is a non-trivial set.

(Small) discrete monoidal categories are the same as monoid structures on the set of objects...

Amar Hadzihasanovic (Jul 01 2025 at 08:58):

So there is plenty of them

John Baez (Jul 01 2025 at 09:08):

Just in case anyone is having trouble following this conversation: I think Fosco solved @Ignat Insarov's original problem of defining a nice kind of generalized monad that looks like $T: \mathcal{C}^J \to \mathcal{C}$.

fosco said:

if $J$ is a monoid, on the other hand, raising to $J$ is a comonad, and $T : C^J \to C$ is an arrow [indeed an endomorphism] in its coKleisli category.

So in the case of $J$ monoid, the correct axioms for your $T$ to be a monad-like gadget are precisely the monad axioms for the coKleisli endo-morphism $T^\sharp : C \rightsquigarrow C$ that T defines.

Ignat Insarov (Jul 01 2025 at 09:25):

Alright! It makes some sense now. But there is still something I do not understand…

Amar Hadzihasanovic said:

$$\mu\colon T \circ (T^J) \circ \mathcal{C}^{-\otimes-} \Rightarrow T$$

What is this $\mathcal C^{-⊗-}$ here? Does it go like $\mathcal C^J \overset {\mathcal C^{-⊗-}} \longrightarrow \mathcal (C^J)^J$?

Amar Hadzihasanovic said:

$$\eta\colon \mathcal{C}^i \Rightarrow T$$

And what is the index $i$ doing here? Is it $i ∈ J$? Do we have $J$ different $η$'s then?

Amar Hadzihasanovic (Jul 01 2025 at 12:45):

Yes, sorry: I am seing $- \otimes -$ as a functor $J \times J \to J$, so that $\mathcal{C}^{- \otimes -}$ goes from $\mathcal{C}^J$ to $\mathcal{C}^{J \times J} \simeq (\mathcal{C}^J)^J$; and $i$ as the functor $1 \to J$ picking the unit object $i$ (here $1$ is the terminal category), so $\mathcal{C}^i$ goes from $\mathcal{C}^J$ to $\mathcal{C}^1 \simeq \mathcal{C}$.

Ignat Insarov (Jul 01 2025 at 13:01):

Alright, this makes sense syntactically! I need some time to figure out if it does what I want, though. In my example of the product co-monad (that I guess was so messed up that no one could understand what I tried to say), I figured I need two different $η$'s, and my index set $J = \{{\rm left};{\rm right}\}$ is not equipped with any monoidal structure.

Jean-Baptiste Vienney (Jul 07 2025 at 03:03):

Maybe, maybe, maybe you can find some interest in this presentation with regard to your question. I'm not sure but I'll post it in case! (To start with, I'm not even sure to understand what you want. What "a functor $T:\mathcal{C}^J \rightarrow \mathcal{C}$ which is a monad in every $j \in J$ is supposed to mean precisely?")

For every $\mathbf{Set}$-operad $\mathcal{O}$, a notion that is called a "$\mathcal{O}$-category" is defined. This a category with a bunch of functors $\mathcal{C}^n \rightarrow \mathcal{C}$ for various integers $n$, multiplication operations and a unit operations. The various integers $n$ depend on the choice of the operad $\mathcal{O}$

For some choice of $\mathcal{O}$ it gives exactly a monad on $\mathcal{C}$. For some other choice of $\mathcal{O}$ it gives exactly a structure of lax monoidal category on $\mathcal{C}$. And of course for other operads, it gives yet other categorical structures.

And then you can also define algebras over a $\mathcal{O}$-category and you recover algebras over a monad and monoid in a lax monoidal category for some choices of $\mathcal{O}$.

Jean-Baptiste Vienney (Jul 07 2025 at 03:05):

Maybe you can specialize to the cases where there is just one functor $\mathcal{C}^n \rightarrow \mathcal{C}$ for some choice of operad $\mathcal{O}$ and maybe you can generalize from $n$ to an arbitrary set $I$ by tweaking the definition to get something as you want, I don't know.

Jean-Baptiste Vienney (Jul 07 2025 at 03:07):

I don't even remember precisely this idea but hopefully I wrote something about it at the time when it was in my head!

Nathaniel Virgo (Jul 07 2025 at 11:23):

There is an old discussion relating Jean-Baptiste Vienney's $\mathcal{O}$-categories at #theory: category theory > (ID my structure) operad in a monoidal bicategory, some of which might or might not be relevant, I don't know.

John Baez (Jul 07 2025 at 12:15):

My old work with James Dolan on coherent O-algebras is also somewhat relevant.

Ignat Insarov (Jul 07 2025 at 13:25):

It will take me some time to look into all these things…