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Stream: theory: category theory

Topic: (ID my structure) operad in a monoidal bicategory

Nathaniel Virgo (Jul 21 2024 at 02:23):

The following seems like a reasonable generalisation of an operad. It could also be seen as a weaker version of a monoid. I'm sure it will be well studied - does anyone know its name, and/or a place where the definition is written out carefully?

Let $\mathcal{C}$ be a monoidal bicategory. Then a ____ in $\mathcal{C}$ is

• an object $C\in\mathcal{C}$
• for each $n\in \{1, 2, \dots\}$, a 1-cell $P_n:C^{\otimes n}\to C$
• a 2-cell ${id}_C\Rightarrow P_1$
• for positive integers $n, k_1, \dots, k_n$, a 2-cell

$$(P_{k_1}\otimes \dots\otimes P_{k_n})\mathrel{;}P_n \Rightarrow P_{k_1 + \dots + k_n}$$

• subject to coherence laws resembling those for an operad.

I'm happy if it turns out to be best defined in a monoidal double category or something, rather than a monoidal bicategory - I was just trying to choose the simplest structure where it seems like it would make sense.

An operad ought to be an example of this, since I think there is a monoidal bicategory where

• objects are natural numbers
• 1-cells are sets
• 2-cells are functions
• composition of 1-cells is cartesian product of sets
• monoidal product is addition on objects and cartesian product of sets on 1-cells

Notification Bot (Jul 21 2024 at 02:28):

This topic was moved here from #learning: questions > (ID my structure) operad in a monoidal bicategory by Nathaniel Virgo.

Rémy Tuyéras (Jul 22 2024 at 00:31):

This seems to intersect with @Jean-Baptiste Vienney's notion of $\mathbb{I}$-ad (see the discussions on his personal space where I think so far he has illustrated the concept with $\mathcal{C} = \mathbf{Cat}$ (categories, functors, natural transformations)). There is also quite a bit of discussion on how the definition you proposed above can be seen as an algebra over some operad too

Nathaniel Virgo (Jul 22 2024 at 03:49):

Cool, that looks very similar, indeed.

Since it's something that's being discussed currently, let me say the thoughts I had about it, since it might connect to the other discussion.

I was thinking about this because I was trying to formulate an alternative definition of virtual equipment, so that's what I'll work towards below.

All of this is tentative - I haven't checked it carefully. The status of it is "hey look at this neat thing I noticed" - it's not something I'm working on seriously, so I don't plan to write it up.

First, I think we can get coloured operads out of this definition, by instantiating it in $\bf Span$. Then the chosen object $C$ becomes the set of types/colours. A 1-cell $P_n:C^{\otimes n}\to C$ gives us the set of morphisms for each type, and the chosen 2-cells give us the composition and identity rules.

Second, I think we can get what we might call a "virtual bicategory", meaning a virtual equipment in which all of the maps in the "tight" (aka vertical) direction are identities. If you think in terms of string diagrams then this is like a coloured operad but where you colour in the gaps between the wires as well as the wires. To do this, given a set $C_0$, consider the following slightly weird looking monoidal bicategory:

• objects are spans $C_0\leftarrow X \to C_0$
• 1-cells are spans of spans. This means a morphism $X\to Y$ is a span $C_0\leftarrow S \to C_0$ together with maps-of-spans $X\leftarrow S \to Y$
• 2-cells are maps of spans-of-spans, which means a function $S\to S'$ such that everything in sight commutes.
• the monoidal product is by composition of spans. So we tensor objects by composing the spans $C_0\leftarrow X \to C_0\leftarrow Y \to C_0$ by pullback, and similarly we tensor 1-cells by composing the spans $C_0\leftarrow S \to C_0\leftarrow T \to C_0$ by pullback
• composition of 1-cells is also composition of spans, but in the other direction: we compose $X\leftarrow S \to Y\leftarrow T \to Z$ by pullback to get a 1-cell $X\to Z$.

Then $C_0$ becomes the set of objects in our virtual bicategory $\mathbb{C}$, and the chosen span $C$ gives us the set of (loose/horiztonal) morphisms between each pair of elements of $\mathbb{C}$. The $P_n$ give us the set of cells of each type, where a type now consists of a sequence of $n$ composable morphisms, together with another morphism whose source and target match those of the sequence. The rest of the definition gives us the composition rule and also gives us a unit cell for each object.

Finally - this one I'm less certain of, but it's the reason I was thinking about this - I think we can get virtual equipments themselves, by instantiating it in the following monoidal bicategory (which I have not proved is actually a monoidal bicategory, but I think it is):

First, fix a category $\mathcal{C}_0$. Then

• objects are profunctors $\mathcal{C}_0\nrightarrow \mathcal{C}_0$
• 1-cells are spans of profunctors, i.e. spans in the category of profunctors $\mathcal{C}_0\nrightarrow \mathcal{C}_0$. This means a morphism $X\to Y$ is a profunctor $S:\mathcal{C}_0\nrightarrow \mathcal{C}_0$ together with natural transformations $X\Leftarrow S \Rightarrow Y$
• 2-cells are maps of spans-of-profunctors, which means a natural transformation $S\to S'$ such that everything in sight commutes.
• the monoidal product is by composition of profunctors. So we tensor objects by composing profunctors $\mathcal{C}_0\nrightarrow \mathcal{C}_0$ by a coend, and similarly we tensor 1-cells by composing profunctors $\mathcal{C}_0\nrightarrow \mathcal{C}_0$ by a coend
• composition of 1-cells is by composition of spans: we compose $X\Leftarrow S \Rightarrow Y\Leftarrow T \Rightarrow Z$ by taking a pullback in the functor category $[\mathcal{C}_0^{op}\times\mathcal{C},\mathbf{Set}]$.

I guess there is work to do to prove that this gives a virtual equipment (if indeed it does). But the idea is that the category $\mathcal{C}_0$ gives us the objects and 'tight' maps in a virtual equipment $\mathbb{C}$. The chosen object $C$ is a profunctor $\mathcal{C}_0\nrightarrow\mathcal{C}_0$, which gives us the set of 'loose' arrows between each pair of objects, and in addition also gives us an action of $\mathcal{C}_0$ on the loose arrows, from the left and the right. The rest of the definition should take care of the units and the composition rule, with everything being compatible with the action. My intuition says that this, or something like it, ought to give an equivalent definition of virtual equipments, although it is just an intuition really.

Jean-Baptiste Vienney (Jul 22 2024 at 12:19):

Let me say that compared to what I was looking at, you replace $\mathbf{Cat}$ by any monoidal bicategory. And compared to your definition, I replace $\mathbb{N}$ by any operad. I ask for a functor $T^t:C^n \rightarrow C$ for every $t \in \mathcal{O}(n)$ for some ($\mathbf{Set}$-based) operad $\mathcal{O}$.

If you define $\mathcal{O}(n)=\{n\}$, $\mathcal{id}=1$ and $n \circ (k_1,…,k_n) = k_1+…+k_n$, you recover your definition in the case $\mathbf{C}=\mathbf{Cat}$. But other choices of $\mathcal{O}$ let you restrict the arities of the functors of the type $C^n \rightarrow C$. The main motivation for using an operad $\mathcal{O}$ is that if you chose $\mathcal{O}(1)=\{\mathrm{id}\}$, and $\mathcal{O}(n)=\emptyset$ then the definition becomes the one of a monad in $C$.

So now, my preferred concept would be by blending these two features. It would give “monoidal monad in a monoidal bicategory” as a special case I guess.

Jean-Baptiste Vienney (Jul 22 2024 at 12:22):

(I don’t find traces of such a name on Google, but I guess this is the natural name.)

Jean-Baptiste Vienney (Jul 22 2024 at 12:31):

But maybe your definition already contains mine in some way, I’m not sure.

John Baez (Jul 22 2024 at 12:39):

A [[monoidal monad]] in the 2-category $\mathbf{Cat}$ is a well-known concept, and it's fairly easy to copy definition in any monoidal bicategory. Someone like Lack or Street should have done it already! But I don't know if they have.

John Baez (Jul 22 2024 at 12:45):

They would say: if $\mathbf{C}$ is a monoidal bicategory, there is a bicategory $\mathbf{PsMon}(\mathbf{C})$ of pseudomonoids in $\mathbf{C}$, and a monad in that bicategory is a monoidal monad in $\mathbf{C}$. This is why we don't let Australia touch the other continents.

Nathaniel Virgo (Jul 22 2024 at 14:06):

How do monoidal monads relate to this? I'm fairly familiar with them but don't immediately see the connection.

John Baez (Jul 22 2024 at 14:09):

Just to be clear, I don't understand that at all; I was merely responding to @Jean-Baptiste Vienney's comment

It would give “monoidal monad in a monoidal bicategory” as a special case I guess. (I don’t find traces of such a name on Google, but I guess this is the natural name.)

Jean-Baptiste Vienney (Jul 22 2024 at 14:09):

I'm no longer sure that monoidal monads relate to what your define. But monads in $\mathbf{Cat}$ at least do. If you chose $\mathcal{C}=\mathbf{Cat}$ and replace $\mathbb{N}$ by any $\mathbf{Set}$-based operad $\mathcal{O}$ as in my definition, and choose the operad $\mathcal{O}$ defined by $\mathcal{O}(1)=\{\mathrm{id}\}$ and $\mathcal{O}(n)=\emptyset$ for $n \neq 1$ then $C$ is a category together with a monad on this category.

Jean-Baptiste Vienney (Jul 22 2024 at 14:13):

That's the only rigorous thing I can say for now.

Jean-Baptiste Vienney (Jul 22 2024 at 14:13):

I can recall my definition to make this clearer.

Jean-Baptiste Vienney (Jul 22 2024 at 14:14):

Just wait one minute and I copy it here.

Nathaniel Virgo (Jul 22 2024 at 14:15):

Thanks! I was just trying to look it up

Jean-Baptiste Vienney (Jul 22 2024 at 14:18):

Let $\mathcal{C}$ be a category and $\mathcal{O}$ an operad. A structure of $\mathcal{O}$-category on $\mathcal{C}$ is given by a functor
$T^t:\mathcal{C}^{n} \rightarrow \mathcal{C}$
for every $n \in \mathbb{N}$ and $t \in \mathcal{O}(n)$, together with:

• natural transformations
  $m_{t,s_1,...,s_{n(t)}}:T^t \circ (T^{s_1} \times ... \times T^{s_n}) \Rightarrow T^{t \circ (s_1,...,s_n)}$

• a natural transformation
  $u:1_\mathcal{C} \Rightarrow T^\mathrm{id}$

such that some associativity and unitality diagrams commute.

Nathaniel Virgo (Jul 22 2024 at 14:31):

Ok cool. So then the combined definition would be

Let $\mathcal{M}$ be a monoidal bicategory and $\mathcal{O}$ an operad. Then a ____ (I'm bad at names) is

• an object $C\in\mathcal{M}$
• for each $t\in\mathcal{O}$ a 1-cell $T^t:C^{\otimes n(t)}\to C$
• for each $t, s_1, \dots, s_{n(t)}\in\mathcal{O}$, a 2-cell

$$(T^{s_1}\otimes\dots\otimes T^{s_{n(t)}})\mathrel{;}T^{t} \Rightarrow T^{(s_1,\dots s_{n(t)})\mathrel{;} t}$$

• a 2-cell $1_C\Rightarrow T^{id}$
• subject to [wave hands]

(where $t\in \mathcal{O}$ means it's one of the morphisms and $n(t)$ is the arity of $t$.)

I like it.

Jean-Baptiste Vienney (Jul 22 2024 at 14:33):

I like it too!

Nathaniel Virgo (Jul 22 2024 at 14:40):

Jean-Baptiste Vienney said:

I'm no longer sure that monoidal monads relate to what your define. But monads in $\mathbf{Cat}$ at least do. If you chose $\mathcal{C}=\mathbf{Cat}$ and replace $\mathbb{N}$ by any $\mathbf{Set}$-based operad $\mathcal{O}$ as in my definition, and choose the operad $\mathcal{O}$ defined by $\mathcal{O}(1)=\{\mathrm{id}\}$ and $\mathcal{O}(n)=\emptyset$ for $n \neq 1$ then $C$ is a category together with a monad on this category.

Do monoidal categories and lax monoidal functors form a monoidal bicategory? If so I think we'd get monoidal monads by using that instead of $\bf Cat$.

Jean-Baptiste Vienney (Jul 22 2024 at 14:54):

I don't know but I think we need just a bicategory to get a monad or a monoidal monad.

Maybe we could replace "monoidal bicategory" by something that I would like to call a monoidal $\mathcal{O}$-bicategory.

In the case where $\mathcal{O}(n)=\{n\}$, I would like a monoidal $\mathcal{O}$-bicategory to be a monoidal bicategory. In the case where $\mathcal{O}(1)=\{\mathrm{id}\}$ and $\mathcal{O}(n)=\emptyset$ if $n \neq 1$, a monoidal $\mathcal{O}$-bicategory to be a bicategory.

I'm not completely clear about that.

Jean-Baptiste Vienney (Jul 22 2024 at 14:56):

The idea is would be that you just need to restrict yourself to a monoidal structure with arities given by $\mathcal{O}$ in your bicategory in order to define a ____ in this bicategory.

Nathaniel Virgo (Jul 22 2024 at 14:58):

hmm, then what would an $\mathcal{O}$-monoidal bicategory be like for a general operad $\mathcal{O}$ (i.e. not a subset of $\N$)? Sounds like it might be a cool thing in itself if it can be defined (but maybe it cant'?).

Jean-Baptiste Vienney (Jul 22 2024 at 15:04):

What definition of a monoidal bicategory do you use? I don't know much about them and the nLab gives several references containing different definitions.

Jean-Baptiste Vienney (Jul 22 2024 at 15:36):

In Higher-dimensional algebra I: braided monoidal 2-categories, Lemma 3, they say that a semistrict monoidal 2-category is a 2-category $\mathcal{C}$ together with $2$-functors $\otimes:\mathcal{C} \otimes_G \mathcal{C} \rightarrow \mathcal{C}$ and $i:\mathcal{I} \rightarrow \mathcal{C}$ such that the usual three diagrams for a monoid in a monoidal category commute, where $\otimes_G$ is the Gray tensor product.

I guess that to obtain semistrict $\mathcal{O}$-monoidal 2-category, you would need something like the Gray tensor product but of the form
$\otimes_{G,t}(\mathcal{C}_1,...,\mathcal{C}_n)$ for every $t \in \mathcal{O}$.

Jean-Baptiste Vienney (Jul 22 2024 at 15:37):

And then ask for $2$-functors $\otimes_{G,t}(\mathcal{C},...,\mathcal{C}) \rightarrow \mathcal{C}$.

Nathaniel Virgo (Jul 22 2024 at 15:38):

I haven't been super precise about that - I don't know what kind of monoidal bicategory would work best. I have a vague feeling it might be one of those things that work better in a (monoidal) double category, but I'm not sure.

Jean-Baptiste Vienney (Jul 22 2024 at 15:50):

I have an idea.

• You start with the definition of an $\mathcal{O}$-category that I gave.

• Then you can define an $\mathcal{O}$-algebra in an $\mathcal{O}$-category $\mathcal{C}$ as an object $A \in \mathcal{C}$ together with a morphism $a_t:T^t(A,...,A) \rightarrow A$ for every $t \in \mathcal{O}(n)$ (where you have $n$ $A$'s) and a morphism $A \rightarrow T^{\mathrm{id}}(A)$ such that this diagram commutes:
  .
  (replace $n(t)$ by $n$)

  

• Maybe the category of $2$-categories and $2$-functors is an $\mathcal{O}$-category for any $\mathcal{O}$ (by starting from the fact that the category of $2$-categories and $2$-functor equipped with the Gray tensor product is a monoidal category).

• An $\mathcal{O}$-monoidal bicategory would be an $\mathcal{O}$-algebra in this $\mathcal{O}$-category.
• Then a ____ should be defined as you wrote before but in arbitrary $\mathcal{O}$-monoidal bicategory.

Jean-Baptiste Vienney (Jul 22 2024 at 15:58):

(by the way, it's maybe easier to look at these slides for the first two points)

Jean-Baptiste Vienney (Jul 22 2024 at 15:59):

So the remaining work to do is the third point (and the fourth).

Jean-Baptiste Vienney (Jul 22 2024 at 16:01):

And then a lot of things should be an ____ ahah.

Jean-Baptiste Vienney (Jul 22 2024 at 16:04):

(apparently, at least the notions of colored operad, monad in a $2$-category, and lax monoidal category (in a monoidal bicategory?))

Rémy Tuyéras (Jul 22 2024 at 20:51):

@Jean-Baptiste Vienney looking at where the conversation is going, maybe it might be an appropriate time to look at my suggestion (I copy-pasted part of it below):

I was looking at them and was wondering if it would be interesting to you to generalize your notion of $\mathcal{O}$-category to a context where you have two operations $\otimes_1$ and $\otimes_2$ replacing $\circ$ and $\times$?

The reason why I say that is that you have morphisms:

$\mathcal{O}(n) \times \mathcal{O}(k_1) \times \dots \times \mathcal{O}(k_n) \to \mathcal{O}(n \circ (k_1,\dots,k_n))$

$T^n \circ (T^{k_1} \times \dots \times T^{k_n}) \to T^{n \circ (k_1+\dots+k_n)}$

Here, the very first "monoidal products" $\circ$ and $\times$ (in the previous morphisms) seem to have a very specific role, hence my wondering about these things:

$X^n \otimes_1 (X^{k_1} \otimes_2 \dots \otimes_2 X^{k_n}) \to X^{n \odot_1 (k_1\odot_2. \dots \odot_2 k_n)}$

where you have some sort of doubly-indexed object $\mathcal{C}_{n,m}$ with:

• $\otimes_1:\mathcal{C}_{n,1} \times \mathcal{C}_{k,n} \to \mathcal{C}_{k,1}$ and
• $\otimes_2:\mathcal{C}_{k_1,1} \times \dots \times \mathcal{C}_{k_n,1} \to \mathcal{C}_{k_1+\dots+k_n,n}$

This would also clarify the definition of your $\mathbb{I}$ using $\mathcal{C}_{k,n}$ as the set of all $n$-partitions of $k$, namely tuples $(k_1,\dots,k_n)$ for which $\sum k_i = k$ such that we now have:

• $\otimes_1$ as the addition
• $\otimes_2$ as the product operation (making tuples)

I provided more details below to elaborate on the intuitions expressed above. One important thing to note is that this formalism does not require you to treat $\mathcal{O}$-categories as some kind of algebras, nor to have 2-cells in your structures. Instead, it allows you to consider some kind of lax morphisms between compatible structures, whose definitions are tailored to capturing the axioms of $\mathcal{O}$-categories as expressed in the examples.

Starter: Define a decomposed category (just to give it a name) as an $\mathbb{N}^{\times 2}$-indexed collection of categories $\mathcal{C}_{k,n}$ equipped with:

• functors $\otimes_1:\mathcal{C}_{n,1} \times \mathcal{C}_{k,n} \to \mathcal{C}_{k,1}$;
• functors $\otimes_2:\mathcal{C}_{k_1,1} \times \dots \times \mathcal{C}_{k_n,1} \to \mathcal{C}_{k_1+\dots+k_n,n}$ and
• a distinguished object $I:\mathbf{1} \to \mathcal{C}_{1,1}$

such that the functor

$\otimes_1 \circ (\mathsf{id} \times \otimes_2): \mathcal{C}_{n,1} \times \mathcal{C}_{k_1,1} \times \dots \times \mathcal{C}_{k_n,1} \to \mathcal{C}_{k,1}$

makes certain diagrams commute (to be determined later)

Step 1: To retrieve the indexing structure $\mathbb{I}$:

• take the discrete categories $\mathcal{C}_{k,n} = \mathbb{I}(k,n) = \{(k_1,\dots,k_n)~|~\sum_ik_i = k\}$ (this is the set of $n$-partitions of $k$);
• define $\otimes_1:(n) \times (k_1,\dots,k_n) \to (k_1+\dots+k_n)$;
• define $\otimes_2:(k_1,\dots,k_n)\mapsto (k_1,\dots,k_n)$ and
• take $I = (1) \in \mathcal{C}_{1,1}$

Step2: To retrieve operads:

• take the categories $\mathcal{C}_{k,n} = \mathbf{Set}^{\otimes n}$;
• define $\otimes_1:(A,B) \mapsto A \times B$;
• define $\otimes_2:(A_1,\dots,A_n) \mapsto A_1 \times \dots \times A_n$ and
• take $I = \{\ast\} \in \mathcal{C}_{1,1}$

Step 3: To retrieve the notions discussed above:

• take the categories $\mathcal{C}_{k,n} = [C^{\otimes k},C^{\otimes n}]$;
• define $\otimes_1:(T,S) \mapsto T \circ S$;
• define $\otimes_2:(T_1,\dots,T_n) \mapsto T_1 \otimes \dots \otimes T_n$ and
• take $I = \mathsf{id}_C \in \mathcal{C}_{1,1}$

Step 4: Denote the free decomposed category generated by the terminal categories $\mathcal{C}_{k,1} = \{\ast\}$ as $(\mathbb{T},\otimes_1,\otimes_2)$

Step 5: Define a morphism $F:(\mathcal{C},\otimes_1,\otimes_2,I_{\mathcal{C}}) \to (\mathcal{D},\otimes_1,\otimes_2,I_{\mathcal{D}})$ between decomposed categories as some kind of lax morphism as follows:

• a collection $F_{k,n}:\mathcal{C}_{k,n} \to \mathcal{D}_{k,n}$ of functors, with
• a morphism $I_{\mathcal{D}} \to F_{1,1}(I_{\mathcal{C}})$ and
• a natural transformation:

$F_{n,1}(X) \otimes_1 (F_{k_1,1}(Y_1) \otimes_2 \dots \otimes_2 F_{k_n,1}(Y_n)) \to F_{k_1+\dots+k_n,1}(X \otimes_1 (Y_1 \otimes_2 \dots \otimes_2 Y_n))$

for which certain diagrams commute. Denote the resulting category as $\mathbf{Dec}$.

Final step: Show that:

• there is a morphism $(\mathbb{T},\otimes_1,\otimes_2) \to (\mathbb{I},\otimes_1,\otimes_2)$ in $\mathbf{Dec}$ (this morphism gives you the arities)

• an operad is a morphism of the form

  $(\mathbb{I},\otimes_1,\otimes_2) \to (\mathcal{C},\otimes_1,\otimes_2)$

  in $(\mathbb{T},\otimes_1,\otimes_2) \downarrow \mathbf{Dec}$, where $\mathcal{C}$ is as described in step 2

• the structures discussed above are morphisms of the form

  $(\mathbb{I},\otimes_1,\otimes_2) \to (\mathcal{C},\otimes_1,\otimes_2)$

  in $(\mathbb{T},\otimes_1,\otimes_2) \downarrow \mathbf{Dec}$, where $\mathcal{C}$ is as described in step 3

To finish: Even if you do not like the concepts above and want to keep looking at other ones, I suspect that the concept of decomposed categories will help you clarify your investigations in other structures because they give you an (almost) minimal presentation of the examples for $\mathcal{O}$-categories

John Baez (Jul 22 2024 at 21:17):

Nathaniel Virgo said:

Jean-Baptiste Vienney said:

I'm no longer sure that monoidal monads relate to what your define. But monads in $\mathbf{Cat}$ at least do. If you chose $\mathcal{C}=\mathbf{Cat}$ and replace $\mathbb{N}$ by any $\mathbf{Set}$-based operad $\mathcal{O}$ as in my definition, and choose the operad $\mathcal{O}$ defined by $\mathcal{O}(1)=\{\mathrm{id}\}$ and $\mathcal{O}(n)=\emptyset$ for $n \neq 1$ then $C$ is a category together with a monad on this category.

Do monoidal categories and lax monoidal functors form a monoidal bicategory? If so I think we'd get monoidal monads by using that instead of $\bf Cat$.

Yes, a monoidal monad is a monad in the bicategory of monoidal categories, lax monoidal functors and natural transformations. There must be other definitions, but they're too complicated for me to remember since I don't think about monoidal monads very often.

We don't need a monoidal bicategory of monoidal categories for this definition to parse. There is, in fact, a monoidal bicategory of monoidal categories where the tensor product of monoidal categories $M$ and $N$ is $M \times N$. But defining the tensor product in $M \times N$ requires the fact that $\mathbf{Cat}$ is a braided monoidal bicategory: we are braiding some letters around each other when we write

$(m,n) \otimes (m', n') = (m \otimes m', n \otimes n')$

So we're seeing an instance of this general pattern: you can define a bicategory $\mathbf{PsMon}(\mathbf{C})$ of monoidal category objects in any monoidal bicategory $\mathbf{C}$, and then "monoidal monads" in $\mathbf{C}$ are monads in $\mathbf{PsMon}(\mathbf{C})$.

But if you want $\mathbf{PsMon}(\mathbf{C})$ to be a monoidal bicategory for some reason, then you should demand the $\mathbf{C}$ be braided. And if you want $\mathbf{PsMon}(\mathbf{C})$ to be braided... well, the story goes on but I don't think you want to hear it right now!

Rémy Tuyéras (Jul 22 2024 at 22:51):

Rémy Tuyéras said:

To finish: Even if you do not like the concepts above and want to keep looking at other ones, I suspect that the concept of decomposed categories will help you clarify your investigations in other structures because they give you an (almost) minimal presentation of the examples for $\mathcal{O}$-categories

Also, with the formalism I gave you, an algebra is given by a certain element $A^k \in \mathcal{C}_{0,k}$ for every $k$. Here, the map

$\otimes_2:\mathcal{C}_{0,1} \times \dots \times \mathcal{C}_{0,1} \to \mathcal{C}_{0,n}$

gives us a correspondence $A^n = \otimes_2 (A^1,\dots,A^1)$ and the map $\otimes_1:\mathcal{C}_{n,1} \times \mathcal{C}_{0,n} \to \mathcal{C}_{0,1}$ gives us an interpretation for $T(A,\dots,A)$.

Specifically, let us denote by $(\mathbb{A},\otimes_1,\otimes_2)$ the free decomposed category generated by the terminal categories $\mathcal{C}_{0,k} = \{\ast\}$.

Conjecture: An algebra is a morphism $(\mathbb{I},\otimes_1,\otimes_2) \to (\mathcal{C},\otimes_1,\otimes_2)$ in the double-coslice category under $(\mathbb{T},\otimes_1,\otimes_2)$ and $(\mathbb{A},\otimes_1,\otimes_2)$.

Rémy Tuyéras (Jul 23 2024 at 00:24):

If the conjecture above is true, then studying quotients of algebras could potentially be done:

• in the same locally presentable category
• by using transfinite pushout constructions (see [[small object argument]]) in a category that mixes both theories and algebras
• could relate any types of algebras (from operads to monads, and so on) so that a quotient or a completion of an algebra may end up being that of another monad or operad
• you may be able to get rid of certain required conditions in your proposition for detecting quotients of algebras as now you would be able to compute colimits in a locally presentable category

Nathaniel Virgo (Jul 23 2024 at 03:22):

Rémy Tuyéras [said]

Starter: Define a decomposed category (just to give it a name) as an $\mathbb{N}^{\times 2}$-indexed collection of categories $\mathcal{C}_{k,n}$ equipped with:

• functors $\otimes_1:\mathcal{C}_{n,1} \times \mathcal{C}_{k,n} \to \mathcal{C}_{k,1}$;
• functors $\otimes_2:\mathcal{C}_{k_1,1} \times \dots \times \mathcal{C}_{k_n,1} \to \mathcal{C}_{k_1+\dots+k_n,n}$ and
• a distinguished object $I:\mathbf{1} \to \mathcal{C}_{1,1}$

such that the functor

$\otimes_1 \circ (\mathsf{id} \times \otimes_2): \mathcal{C}_{n,1} \times \mathcal{C}_{k_1,1} \times \dots \times \mathcal{C}_{k_n,1} \to \mathcal{C}_{k,1}$

makes certain diagrams commute (to be determined later)

I sort of feel like this really wants to be just an $\mathbb{N}$-indexed family of categories with a functor $\mathcal{C}_n\times \mathcal{C}_{k_1}\times \dots\times \mathcal{C}_{k_n}\to \mathcal{C}_{k_1+\dots+k_n}$ and an object of $\mathcal{C}_1$, making diagrams commute. (So just define it in terms of $\otimes_1 \circ (\mathsf{id} \times \otimes_2)$ directly instead of via $\otimes_1$ and $\otimes_2$.) Which then seems like either an operad in $\mathbf{Cat}$ or some kind of lax version of that, depending on what the diagrams are. (Surely this is a well known thing!) Or am I barking up the wrong tree?

Nathaniel Virgo (Jul 23 2024 at 03:29):

On another note, can we think of any good examples of $\mathcal{O}$-monoidal categories for an operad $\mathcal{O}$ that's not a set of arities? It might help pump the intuition.

By an $\mathcal{O}$-monoidal category I mean a category equipped with a functor $\otimes_t:\mathcal{C}^{\times n(t)}\to\mathcal{C}$ for each $t\in \mathcal{O}$, such that (i) $\otimes_{1} = id_\mathcal{C}$ and (ii)

image.png

commutes for each $s, t_1, \dots, t_{n(s)}\in \mathcal{O}$. That seems an easier place to start than $\mathcal{O}$-monoidal bicategories.

I guess I'm looking for an example where we have $n(t)=n(t')$ but not $\otimes_t = \otimes_{t'}$, for some $t,t'\in\mathcal{O}$.

Jean-Baptiste Vienney (Jul 23 2024 at 04:09):

@Rémy Tuyéras I have some trouble understanding your suggestion. Maybe it’s too general for me and I can’t relate it to structures I know. I’m also interested in the same question that @Nathaniel Virgo. What is an operad such that some $\mathcal{O}(n)$ is of cardinal $\ge 2$? I think a “set of arities” is just an operad with $\le 1$ elements in each $\mathcal{O}(n)$?

Jean-Baptiste Vienney (Jul 23 2024 at 04:11):

Hmm, maybe part of your suggestion is that the tensor product takes different categories in each entry, right? It looks like a cool idea.

Jean-Baptiste Vienney (Jul 23 2024 at 04:13):

Maybe it would help obtaining structures with “multiple types”.

Nathaniel Virgo (Jul 23 2024 at 04:17):

Right, that's what I meant by "set of arities". The main examples of operads that I know are the various "operads of wiring diagrams", such as example 1 in Libkind's An Algebra of Resource Sharing Machines, which has more than one element in most of the $\mathcal{O}(n)$'s.

Each $\otimes_t$ takes in $n(t)$ copies of $\mathcal{C}$ and combines them into 1. It's meant to be a generalisation of an unbiased monoidal category, where instead of having a monoidal product of $n$ objects for every $n\in\mathbb{N}$, we have a monoidal product of $n(t)$ objects for every $t\in\mathcal{O}$. I think this is the kind of thing you were proposing earlier - is that right?

Nathaniel Virgo (Jul 23 2024 at 04:18):

Although, I realised that strict monoidal categories are an example of $\mathcal{O}$-monoidal categories according to my definition, but ordinary (weak) monoidal categories aren't, so there's probably something missing.

Nathaniel Virgo (Jul 23 2024 at 04:20):

(I'm not sure if we really need this btw, I'm just curious about it.)

Rémy Tuyéras (Jul 23 2024 at 06:02):

Nathaniel Virgo said:

I sort of feel like this really wants to be just an $\mathbb{N}$-indexed family of categories with a functor $\mathcal{C}_n\times \mathcal{C}_{k_1}\times \dots\times \mathcal{C}_{k_n}\to \mathcal{C}_{k_1+\dots+k_n}$ and an object of $\mathcal{C}_1$, making diagrams commute. (So just define it in terms of $\otimes_1 \circ (\mathsf{id} \times \otimes_2)$ directly instead of via $\otimes_1$ and $\otimes_2$.) Which then seems like either an operad in $\mathbf{Cat}$ or some kind of lax version of that, depending on what the diagrams are. (Surely this is a well known thing!) Or am I barking up the wrong tree?

If I did that, I do not know how I would get the definition for algebras within the same language (algebras use $\mathcal{C}_{0,n}$ while the restriction proposed above only looks at $\mathcal{C}_{n,1}$, which means that an algebra would need to be of the form $T_n(A) \to A$ and not $T_n(A,\dots,A) \to A$).

Specifically, the definition I provide above should allow me to embed algebras and theories (monads, operads, etc.) in a category where I can construct the former from the latter in the same language. The hope with this is that I should be able to answer questions such as "if my monad does this, does that mean my algebras do that?" more easily.

Rémy Tuyéras (Jul 23 2024 at 06:59):

After reflection, the algebra side of the model might actually be possible with what you proposed @Nathaniel Virgo

Nayan Rajesh (Jul 23 2024 at 09:33):

I might be a little late to the discussion, but this seems like Day and Street's notion of a 'lax monoid' in a Gray monoid ($\cong$ monoidal bicategory) . Here is a screenshot of their (unwrapped) definition:

image.png

Indeed, initiating this definition in Span returns multicategories (=colored operads), and the reference for all of this is "Lax monoids, pseudo-operads and convolution".

Reading the rest of the discussion seems to indicate that you are looking for something a little more complex now, but I hope this helps!

Jean-Baptiste Vienney (Jul 23 2024 at 14:27):

Nathaniel Virgo said:

Although, I realised that strict monoidal categories are an example of $\mathcal{O}$-monoidal categories according to my definition, but ordinary (weak) monoidal categories aren't, so there's probably something missing.

Your definition is almost the definition of “$\mathcal{O}$-categories” in my slides which is the specialization to $\mathcal{C}=\mathbf{Cat}$ of what you wrote a little before. I think ordinary monoidal categories satisfy this definition, what is the issue? The only difference with what we wrote before is that you’re asking for $\otimes_{1}=1_C$ instead of just a natural transformation $1_{C} \rightarrow \otimes_{1}$. It doesn’t change much for (weak) monoidal categories, but it is important for monads (because you don’t want only trivial monads). In the case of a monad, $\otimes_1$ is precisely the endofunctor of the monad.

Jean-Baptiste Vienney (Jul 23 2024 at 14:30):

Nayan Rajesh said:

I might be a little late to the discussion, but this seems like Day and Street's notion of a 'lax monoid' in a Gray monoid ($\cong$ monoidal bicategory) . Here is a screenshot of their (unwrapped) definition:

image.png

Indeed, initiating this definition in Span returns multicategories (=colored operads), and the reference for all of this is "Lax monoids, pseudo-operads and convolution".

Reading the rest of the discussion seems to indicate that you are looking for something a little more complex now, but I hope this helps!

Ok, so @Nathaniel Virgo’s notion was exactly a lax monoid in a Gray monoid. So now it’s clear that what we add to the story must be the $\mathcal{O}$!

Nathaniel Virgo (Jul 23 2024 at 14:36):

Jean-Baptiste Vienney said:

Your definition is almost the definition of “$\mathcal{O}$-categories” in my slides which is the specialization to $\mathcal{C}=\mathbf{Cat}$ of what you wrote a little before. I think ordinary monoidal categories satisfy this definition, what is the issue? The only difference with what we wrote before is that you’re asking for $\otimes_{1}=1_C$ instead of just a natural transformation $1_{C} \rightarrow \otimes_{1}$. It doesn’t change much for (weak) monoidal categories, but it is important for monads (because you don’t want only trivial monads).

Oh, I see, I hadn't quite realised they were the same. It probably should be a natural transformation $1_{C} \rightarrow \otimes_{1}$. So then shouldn't the "$\mathcal{O}$-monoidal bicategories" you wanted be this sort of thing as well? Or is that what you were saying here? Sorry, I'm just catching up.

Jean-Baptiste Vienney (Jul 23 2024 at 15:12):

The definition of “semistrict” $\mathcal{O}$-monoidal 2-category should be a special case of this.

A semistrict monoidal $2$-category is exactly a Gray monoid i.e. a monoid in the monoidal category $\mathbf{Gray}$ of $2$-categories and $2$-functors with the Gray tensor product. A semistrict $\mathcal{O}$-monoidal $2$-category should be a Gray $\mathcal{O}$-monoid i.e. an algebra (in the same sense as in my slides) in some $\mathcal{O}$-monoidal category $\mathcal{O}-\mathbf{Gray}$ of $2$-categories and $2$-functors together with some notion of $\mathcal{O}$-monoidal Gray tensor product of $2$-categories.

But defining this $\mathcal{O}$-monoidal Gray tensor product of $2$-categories, if it works, should take some work given that the usual notion of Gray tensor product isn’t very easy apparently.

Rémy Tuyéras (Jul 24 2024 at 19:25):

Rémy Tuyéras said:

After reflection, the algebra side of the model might actually be possible with what you proposed Nathaniel Virgo

@Nathaniel Virgo @Jean-Baptiste Vienney I would like to take back this last claim (I think I got a bit confused with my own examples too) and still want to put forward that decomposed categories (as defined above with $\otimes_1$ and $\otimes_2$) are relevant to the discussion. I believe it strongly relates to the points that you two were discussing for bicategories and weak monoidal categories, and, in particular, the example with spans.

First, I defined the decomposed category $\mathbb{T}$ as something free on a prestructure $\mathcal{C}_{k,1} = \{\ast\}$. Here, the idea is that the resulting $\mathbb{T}_{k,1}$ will have $\geq 2$ elements. In other words, the generated structure $\mathbb{T}$ will capture weak monads of the form $T_t:C^{t} \to C$ (where $t$ is a non-binary tree and $C^{t}$ is the tensor for the corresponding bracketing), which will be generated out of generators of the form $T_k:C^{k} \to C$. These "generators" are not of the form $T_k:(C^1)^t\to C$ even though that part of the monad will also be captured by one of the generated elements of $\mathbb{T}$. Instead, the generators can address the lax/biased/weak case by requesting an arrow $(C^1)^t \to C^{n(t)}$ as a way to "reorder the brackets" in $(C^1)^t$. I believe this would be equivalent to requiring the image of the generator $\mathbb{T}_{k,1} = \{\ast\}$ to be at least weakly terminal in the decomposed categories it is sent to.

Similarly, you can define weak algebras with the object $\mathbb{A}$. This time you get morphisms $T:A^t \to A$ (where $t$ is a non-binary tree and $A^{t}$ is the tensor for the corresponding bracketing) out of generators of the form $T_k:A^k \to A$. These "generators" are also not of the form $T_k:(A^1)^t \to A$ even though that part of the algebra will also be captured by one of the generated elements of $\mathbb{A}$.

And I believe that this also answers in some way the following point too:

Jean-Baptiste Vienney said:

I’m also interested in the same question that Nathaniel Virgo. What is an operad such that some $\mathcal{O}(n)$ is of cardinal $\ge 2$? I think a “set of arities” is just an operad with $\le 1$ elements in each $\mathcal{O}(n)$?

Jean-Baptiste Vienney (Jul 24 2024 at 20:49):

Rémy Tuyéras said:

$X^n \otimes_1 (X^{k_1} \otimes_2 \dots \otimes_2 X^{k_n}) \to X^{n \odot_1 (k_1\odot_2. \dots \odot_2 k_n)}$

@Rémy Tuyéras, what are these $\odot_1,\odot_2$?

Rémy Tuyéras (Jul 24 2024 at 20:51):

The notation $\odot_1$ and $\odot_2$ were informal in this post, but I formalized them in my Step 1, where I defined $\mathbb{I}$ as a decomposed category

Rémy Tuyéras (Jul 24 2024 at 20:57):

I changed the notations $\odot_1$ and $\odot_2$ to $\otimes_1$ and $\otimes_2$

Jean-Baptiste Vienney (Jul 24 2024 at 21:13):

Ok, thanks.

Jean-Baptiste Vienney (Jul 24 2024 at 21:14):

Rémy Tuyéras said:

Jean-Baptiste Vienney looking at where the conversation is going, maybe it might be an appropriate time to look at my suggestion (I copy-pasted part of it below):

Step 4: Denote the free decomposed category generated by the terminal categories $\mathcal{C}_{k,1} = \{\ast\}$ as $(\mathbb{T},\otimes_1,\otimes_2)$

I don't understand this.

Jean-Baptiste Vienney (Jul 24 2024 at 21:15):

Rémy Tuyéras said:

Final step: Show that:

• there is a morphism $(\mathbb{T},\otimes_1,\otimes_2) \to (\mathbb{I},\otimes_1,\otimes_2)$ in $\mathbf{Dec}$ (this morphism gives you the arities)

And here, I don't understand what is $(\mathbb{I},\otimes_1,\otimes_2)$. Is it obtained from the decomposed category?

Jean-Baptiste Vienney (Jul 24 2024 at 21:18):

Do you define $\mathbb{I}(k,n):=\mathcal{C}_{k,n}$? because you wrote $\mathcal{C}_{k,n}=\mathbb{I}(k,n)$ so it look like you're creating a decomposed category from some $\mathbb{I}$.

Jean-Baptiste Vienney (Jul 24 2024 at 21:21):

Your notion looks interesting. I just need some more help to understand what you're doing.

Rémy Tuyéras (Jul 24 2024 at 21:24):

Jean-Baptiste Vienney said:

Step 4: Denote the free decomposed category generated by the terminal categories $\mathcal{C}_{k,1} = \{\ast\}$ as $(\mathbb{T},\otimes_1,\otimes_2)$

I don't understand this.

To construct $\mathbb{T}$ you need a free construction. Start with the $\mathbb{N}^{\times 2}$-collection of discrete categories defined as follows:

• $\mathcal{C}_{k,1} = \{\ast\}$ if $k>0$;
• $\mathcal{C}_{k,n} = \emptyset$ otherwise.

When you "force" $\mathcal{C}_{k,1}$ to become a decomposed category, you will create the free element $\ast \otimes_2 \ast \in \mathcal{C}_{2,1}$ to create an image for the function $\otimes_2:\mathcal{C}_{1,1} \times \mathcal{C}_{1,1} \to \mathcal{C}_{2,1}$. But you will also have the elements

$\ast \otimes_2 (\ast \otimes_2 \ast)\in \mathcal{C}_{3,1}$

and

$(\ast \otimes_2 \ast) \otimes_2 (\ast \otimes_2 \ast)\in \mathcal{C}_{4,1}$

and so on.

You will do that for all your elements using all combinations of $\otimes_1$ and $\otimes_2$. At the end, you should obtain things like trees.

Rémy Tuyéras (Jul 24 2024 at 21:28):

Jean-Baptiste Vienney said:

And here, I don't understand what is $(\mathbb{I},\otimes_1,\otimes_2)$. Is it obtained from the decomposed category?

Here, $\mathbb{I}$ is not a free structure. It really is what is defined in Step 1. I define it a bit differently from how you defined it in earlier post, but it is the corresponding translation of your $\mathbb{I}$ into decomposed categories.

So take $\mathbb{I}(k,n)$ to be the set of $n$-partitions of the integer $k$.

You can equip this $\mathbb{N}^{\times 2}$-collection with operations $\otimes_1$ and $\otimes_2$ as I defined them in the 1st step.

Jean-Baptiste Vienney (Jul 24 2024 at 22:33):

Rémy Tuyéras said:

• a distinguished object $I:\mathbf{1} \to \mathcal{C}_{1,1}$

Am I supposed to get a unit $A \rightarrow T(A)$ of monad from this (in a very simple case of decomposed category)?

It looks like it just gives me a category. Moreover the monoidal unit in a monoidal category is supposed to come from some $\otimes^0$, which is optional. In this way, you can model both unitary and unitless monoidal categories. The additional thing in addition to the multiplications is really the $A \rightarrow \otimes^1(A)$ which is both in the definition of monad and (lax unbiased) monoidal category.

Jean-Baptiste Vienney (Jul 24 2024 at 22:40):

But maybe I don’t understand at all what is the role of this $I$.

Rémy Tuyéras (Jul 25 2024 at 00:09):

Jean-Baptiste Vienney said:

Rémy Tuyéras said:

• a distinguished object $I:\mathbf{1} \to \mathcal{C}_{1,1}$

Am I supposed to get a unit $A \rightarrow T(A)$ of monad from this (in a very simple case of decomposed category)?

Here, with $I:\mathbf{1} \to \mathcal{C}_{1,1}$, you are supposed to get an identity morphism $\mathsf{id}:C \to C$ and it is the lax morphisms that will give you the unit $\mathsf{id} \to F(\mathsf{id})$.

Jean-Baptiste Vienney (Jul 25 2024 at 00:29):

Is a decomposed category meant to model the “$\mathcal{O}$-monoidal bicategory” and then the $\mathcal{O}$-monoidal category is an algebra for this decomposed category?

Jean-Baptiste Vienney (Jul 25 2024 at 00:31):

Or is a decomposed category meant to model the $\mathcal{O}$-monoidal category directly?

Rémy Tuyéras (Jul 25 2024 at 00:31):

Jean-Baptiste Vienney said:

Or is a decomposed category meant to model the $\mathcal{O}$-monoidal category directly?

Yes

Rémy Tuyéras (Jul 25 2024 at 00:31):

An algebra for decomposed categories should correspond to your algebras

Jean-Baptiste Vienney (Jul 25 2024 at 00:31):

Ok

Jean-Baptiste Vienney (Jul 25 2024 at 00:34):

Rémy Tuyéras said:

Jean-Baptiste Vienney said:

Rémy Tuyéras said:

• a distinguished object $I:\mathbf{1} \to \mathcal{C}_{1,1}$

Am I supposed to get a unit $A \rightarrow T(A)$ of monad from this (in a very simple case of decomposed category)?

Here, with $I:\mathbf{1} \to \mathcal{C}_{1,1}$, you are supposed to get an identity morphism $\mathsf{id}:C \to C$ and it is the lax morphisms that will give you the unit $\mathsf{id} \to F(\mathsf{id})$.

What is this lax morphism? Is it a $\otimes_1$ or a $\otimes_2$?

Rémy Tuyéras (Jul 25 2024 at 00:35):

See Step 5 above (I wanted to quote it but the message is long and I am on my phone)

Jean-Baptiste Vienney (Jul 25 2024 at 00:37):

Ok, I see.

Jean-Baptiste Vienney (Jul 25 2024 at 00:38):

I think I start to get the idea. If you take a very simple $\otimes_1$ or a very simple $\otimes_2$ it should give you something interesting?

Rémy Tuyéras (Jul 25 2024 at 00:46):

Another point of this is that you have theories (operads, monads, etc.) and algebras in the set context. Instead of having: "take a endofunctor" for theories and "take an an object" for the algebras, you have "take a decomposed category" and "take a decomposed category"

Rémy Tuyéras (Jul 25 2024 at 00:49):

It is like an algebra wants to be of the form "$t \to 1$" but a monad has another $t$ layer of the form $tt\to t$. With decomposed categories, this is resolved by the indexing

Jean-Baptiste Vienney (Jul 25 2024 at 00:52):

Ok, so in Step 5, the first decomposed category encode the “shape” of the categorical structure, the second one represents the “environment” in which you interpret categorical structures of this shape and the morphism of decomposed structure (or rather its image) represents the categorical structure. Is it a good interpretation?

Rémy Tuyéras (Jul 25 2024 at 00:53):

Basically, what I am saying is that algebras are supposed to be more decomposed than theories, because they mostly use $\otimes_2$, but theories will want to use both $\otimes_1$ and $\otimes_2$

Jean-Baptiste Vienney (Jul 25 2024 at 01:04):

Hmm, I’ll try to understand this later.

Nathaniel Virgo (Jul 25 2024 at 03:16):

@Rémy Tuyéras I'm trying to follow but I keep getting stuck on the definition:

Rémy Tuyéras said:

Starter: Define a decomposed category (just to give it a name) as an $\mathbb{N}^{\times 2}$-indexed collection of categories $\mathcal{C}_{k,n}$ equipped with:

• functors $\otimes_1:\mathcal{C}_{n,1} \times \mathcal{C}_{k,n} \to \mathcal{C}_{k,1}$;
• functors $\otimes_2:\mathcal{C}_{k_1,1} \times \dots \times \mathcal{C}_{k_n,1} \to \mathcal{C}_{k_1+\dots+k_n,n}$ and
• a distinguished object $I:\mathbf{1} \to \mathcal{C}_{1,1}$

such that the functor

$\otimes_1 \circ (\mathsf{id} \times \otimes_2): \mathcal{C}_{n,1} \times \mathcal{C}_{k_1,1} \times \dots \times \mathcal{C}_{k_n,1} \to \mathcal{C}_{k,1}$

makes certain diagrams commute (to be determined later)

Do you think you could sketch out what the equations should say? It seems really weird if the only conditions are on $\otimes_1 \circ (\mathsf{id} \times \otimes_2)$ and don't involve $\otimes_1$ and $\otimes_2$ individually, and that's making it hard to get an intuition for what kind of thing a decomposed category is meant to be.

Rémy Tuyéras (Jul 25 2024 at 08:52):

Sure! To me, the associativity is not as weird as a distributive law (e.g. duoidal-like axioms, $\mathsf{and}$-$\mathsf{or}$ distributivity, etc.)

$\otimes_1(x, \otimes_2( \otimes_1(x_1, \otimes_2(\overline{y}_1)),\dots, \otimes_1(x_n, \otimes_2(\overline{y}_n)) )) = \otimes_1(\otimes_1(x,\otimes_2(x_1,\dots,x_n)), \otimes_2(\overline{y}_1,\dots,\overline{y}_n))$

And the unity axiom is not as weird as a commutative law:

$\otimes_1(x,\otimes_2(I,\dots,I)) = x = \otimes_2(\otimes_1(I,x))$

To me it feels like this has the same kind of flavor as a distributive law between monads

Nathaniel Virgo (Jul 25 2024 at 09:19):

If we use just sets instead of categories, then one source of examples would be to take a [[PRO]] ( = monoidal category where every object is of the form $X^{\otimes n}$ for a distinguished object $X$) $\mathcal{P}$ and then let $C_{n,k}$ be the hom-set $\mathcal{P}(X^n\to X^k)$. Then define $\otimes_2$ as the monoical product in $\mathcal{P}$ and $\otimes_1$ as composition in $\mathcal{P}$. Presumably you can do the same kind of thing starting with some kind of 2-category version of a PRO to get an instance of your definition. So your definition seems like kind of a weaker version of some kind of 2-PRO, where you're only allowed to tensor morphisms with one output wire, and you're only allowed to compose morphisms if the composite has only one output wire.

But, hmm, with this picture in mind I'm not seeing how this gives me operads. Can you expand on how that works?

Rémy Tuyéras said:

Step2: To retrieve operads:

• take the categories $\mathcal{C}_{k,n} = \mathbf{Set}^{\otimes n}$;
• define $\otimes_1:(A,B) \mapsto A \times B$;
• define $\otimes_2:(A_1,\dots,A_n) \mapsto A_1 \times \dots \times A_n$ and
• take $I = \{\ast\} \in \mathcal{C}_{1,1}$

Rémy Tuyéras (Jul 25 2024 at 09:31):

Operads are given by morphisms of decomposed categories of the form

$F:(\mathbb{I},\otimes_1,\otimes_2) \to (\mathcal{C},\otimes_1,\otimes_2)$

in the coslice category $(\mathbb{T},\otimes_1,\otimes_2)\downarrow \mathbf{Dec}$. The definition is given by Step 5 (see my post of July 23rd -- Tokyo time). Since morphisms of decomposed categories are a kind of lax morphism, this essentially means that for every $n$-partition $(k_1,\dots,k_n)$, we have a set $F(k_1,\dots,k_n)$ equipped with:

• a function $\{\ast\} \to F(1)$
• a function $F(n) \times F(k_1) \times \dots \times F(k_n) \to F(k_1+\dots+k_n)$

such that some axioms hold.

Rémy Tuyéras (Jul 25 2024 at 09:42):

Maybe here I could follow your suggestion and decouple the axioms satisfied by $F$ as follows (without affecting the definition of an operad since $F(k_1,\dots,k_n)$ is almost like a formal object):

$F(X) \otimes_1 F(Y) \to F(X \otimes_1 Y)$

$\otimes_2 (F(Y_1),\dots,F(Y_n)) \to F(\otimes_2 (Y_1,\dots,Y_n))$.

This would give me the maps:

$F(n) \times F(k_1,\dots,k_n) \to F(k_1+\dots+k_n)$

$F(k_1)\times \dots \times F(k_n) \to F(k_1,\dots,k_n)$

Rémy Tuyéras (Jul 26 2024 at 21:56):

Rémy Tuyéras said:

Basically, what I am saying is that algebras are supposed to be more decomposed than theories, because they mostly use $\otimes_2$, but theories will want to use both $\otimes_1$ and $\otimes_2$

I think there is better way to say this, so I will try to clarify the intuition.

When you look at operads and their algebras, they can appear different because you have the operations:

$\odot: O(n) \times O(k_1) \times \dots \times O(k_n) \to O(k_1+\dots+k_n)$

$\odot: O(n) \times A^{\times n} \to A$

but you start understanding that they are the same things if you denote $A(k) = A$ because then you have:

$\odot: O(n) \times A(k_1) \times \dots \times A(k_n) \to A(k_1+\dots+k_n)$

$\odot: O(n) \times O(k_1) \times\dots \times O(k_n) \to O(k_1+\dots+k_n)$

Now, if you wanted to address weak lax cases, you would probably want something that can account for different ways to use brackets, which intuitively would mean this:

$\odot: O(n) \times ((A(k_1) \times A(k_2)) \times \dots \times A(k_n)) ) \to A(((k_1+k_2)+\dots+k_n))$

$\odot: O(n) \times ((O(k_1) \times O(k_2)) \times \dots \times O(k_n)) ) \to O(((k_1+k_2)+\dots+k_n))$

This means that the leftmost input of $\odot$ determines the number of elements $k_i$ and the other inputs determine the bracketing in $((k_1+k_2)+\dots+k_n)$. This suggests two parameterizations of $\odot$, namely a decomposition in terms of $\otimes_1$ and $\otimes_2$, such that we have one operation for bracketing and one operation for
the number of inputs. This allows us to treat the weak case only through $\mathcal{C}_{k,n}$ in much the same way weakness is treated in homotopy theory / higher category theory / (lax) coherence theory : by only focusing on the relationships between the elements in the categories $\mathcal{C}_{k,n}$.

Rémy Tuyéras (Jul 26 2024 at 23:10):

And this leads me to say that the decomposed category $\mathbb{A}$, which is to capture algebras, should be the decomposed category that is freely generated on the prestructure:

• $\mathcal{C}_{n,k} = \{\ast\}$

This is to be compared with the definition of $\mathbb{T}$, which should now be freely generated on (to account for the lax cases):

• $\mathcal{C}_{n,k} = \{\ast\}$ if $n > 0$
• $\mathcal{C}_{n,k} = \emptyset$ otherwise

Note that we have a morphism $\mathbb{T} \to \mathbb{A}$ which allows us to talk about the theory associated with an algebra. This suggests that decomposed categories give you a formalism for algebras mainly, and after doing calculations on them (say quotients), you could say "this quotient thing can be seen as an algebra if we take this associated theory" and then "this is an algebra for a monad if the associated theory can be mapped to a monad"