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Stream: deprecated: id my structure

Topic: Bimonoidal functor

Patrick Nicodemus (Aug 06 2023 at 00:19):

any examples of functors $F : C\times D\to E$ where $C, D,E$ are monoidal categories, which are (lax, colax, strongly) monoidal in one variable when the other is fixed, or vice versa?

Currently thinking about double categories but monoidal categories seem like a simpler place to start

James Deikun (Aug 06 2023 at 00:25):

Have you tried functor application $C^C \times C \to C$ where $C$ is monoidal and $C^C$ has the (corresponding) Day convolution monoidal structure?

James Deikun (Aug 06 2023 at 00:28):

Even better would be $\bold{Mon}(C^C) \times C \to C$ -- that one should be monoidal in both variables.

James Deikun (Aug 06 2023 at 00:30):

(Although maybe you need $C$ to be symmetric monoidal in that case.)

Patrick Nicodemus (Aug 06 2023 at 00:40):

Ok, that's helpful. I know I asked for examples but what I really want is the right setting to work in to think about this specific problem I have, and I was hoping that similar examples would help me find the right level of generality/ some theoretical context.

I have something close to a profunctor between double categories A^op x B -> Sets, but it's not "vertically functorial" in the appropriate sense, it's "vertically functorial" in one variable while the other is held fixed.
That is, if $v : a_0\to a_1, v' : a_1\to a_2, w : b_0\to b_1$ are vertical arrows, I have a map $(v,w)\times P(v',w)\to P(v\times v',w)$ and similarly when the situation is dualized, but not a map $P(v,w)\times P(v',w')\to P(vv',ww')$.

I guess what I'm looking for is like, a way of recasting this problem into another way of looking at it which would not require this ugly notion of "almost a profunctor but not quite"

James Deikun (Aug 06 2023 at 00:53):

I think application $\bold{Mon}(C^C) \times C \to C$ fails to be jointly monoidal in the same way when $C$ is not symmetric, actually? You get lax preservation of the monoidal structure in the second variable from the monoid structure in the first variable, and lax preservation of the monoid structure in the first variable from the properties of Day convolution, but I don't think you can combine them without the symmetry.

James Deikun (Aug 06 2023 at 00:57):

So maybe you're missing a nice assumption similar to symmetry that's needed for good behaviour, or maybe you're looking at something as a profunctor when you should be recurrying it and looking at it as some kind of ordinary functor.

Patrick Nicodemus (Aug 06 2023 at 02:17):

Yeah, that makes sense. I recast the problem using a kind of profunctors-as-spans perspective and here's what I got.

All the following arrows are in Cat, i.e., functors.
I have a "base" double category $\mathbb{C}$, which is a category of objects $\mathbb{C}_o$ and of arrows $\mathbb{C}_a$ together with domain, codomain, identity, composition functors.
Then the "almost profunctor" on $\mathbb{C}$ takes the form of a span in Cat from $\mathbb{C}_{a}^{\rm op}$ to $\mathbb{C}_a$, call it $(L, R) : \mathbb{T} \to \mathbb{C}_a^{\rm op}\times \mathbb{C}_a$,
together with two double category structures on $\mathbb{T}$. In both cases $\mathbb{T}$ is the category of arrows, but the objects are different and the composition structures are different.

• In the first double category structure on $\mathbb{T}$, $\mathbb{C}_o^{\rm op}\times \mathbb{C}_a$ is the category of objects, with $(dom\circ L, R)$ and $(cod\circ L,R)$ as the domain and codomain functors; with this double category structure, $L$ extends to a double functor $\mathbb{T}\to\mathbb{C}^{\rm op}$, (for the behavior of this functor on objects, taking the obvious projection)
• In the second category structure, $\mathbb{C}_a^{\rm op}\times \mathbb{C}_o$ is the category of objects for $\mathbb{T}$, and $(L, dom\circ R)$ and $(L, cod\circ R)$ are the domain and codomain functors, $R$ is a double functor $\mathbb{T}\to \mathbb{C}$.

In this way $\mathbb{T}$ is almost a span from $\mathbb{C}^{\rm op}$ to $\mathbb{C}$ in double categories except that it's not a double category in a consistent way.

James Deikun (Aug 06 2023 at 02:59):

Can you view this as an endoprofunctor on the "double category of vertical arrows" of $\mathbb{C}$? Basically the category of (potentially lax) double functors from the walking vertical arrow to $\mathbb{C}$. That would make the lack of any direct reference to $\mathbb{C}_{o}$ in the definition of $\mathbb{T}$ make sense.

Patrick Nicodemus (Aug 06 2023 at 03:00):

Ok, let me think about that.

James Deikun (Aug 06 2023 at 03:28):

Thinking of these as shapes ... objects of $\mathbb{T}$ look like "H" shapes, with vertical arrows as the verticals of the H and the object of $\mathbb{T}$ providing the crossbar. Arrows of $\mathbb{T}$ look like "girders" with one of the planes a square going backward and one a square going forward, and the arrow of $\mathbb{T}$ providing the crossbar. Composition in $\mathbb{T}$ joins girders H-shaped end to H-shaped end. It would also make sense to potentially have a composition joining the H shapes vertically, but joining them horizontally wouldn't make as much sense since the orientation of the planes on the girders wouldn't match so you couldn't extend it to a composition of girders.

Patrick Nicodemus (Aug 06 2023 at 04:35):

Yeah I agree this picture is a good way to view it.

Patrick Nicodemus (Aug 06 2023 at 05:24):

James Deikun said:

Thinking of these as shapes ... objects of $\mathbb{T}$ look like "H" shapes, with vertical arrows as the verticals of the H and the object of $\mathbb{T}$ providing the crossbar. Arrows of $\mathbb{T}$ look like "girders" with one of the planes a square going backward and one a square going forward, and the arrow of $\mathbb{T}$ providing the crossbar. Composition in $\mathbb{T}$ joins girders H-shaped end to H-shaped end. It would also make sense to potentially have a composition joining the H shapes vertically, but joining them horizontally wouldn't make as much sense since the orientation of the planes on the girders wouldn't match so you couldn't extend it to a composition of girders.

Yeah. If you have two H's, say $A$ and $B$, both with a common left hand side, you can vertically compose them if the codomain of the right hand side of $A$ matches the domain of the RHS of $B$.
Similarly, if $A$ and $B$ have a common right hand side, you can compose them if the codomain of the LHS of $A$ matches the domain of the LHS of $B$.

Patrick Nicodemus (Aug 06 2023 at 05:24):

But the H's cannot be composed "top to bottom" in the naive sense. afaict. They already need to be overlapping on the left to be composed on the right and vice versa.

Patrick Nicodemus (Aug 06 2023 at 05:26):

That's the frustrating part. It would make things much simpler

Morgan Rogers (he/him) (Aug 06 2023 at 10:19):

I have been working with a tensor product of symmetric monoidal categories which has the universal property that functors $\mathbb{C}\times \mathbb{D} \to \mathbb{E}$ which are monoidal in each variable with the other fixed factor uniquely as a monoidal functor out of the tensor product. It's constructed by adding some natural isomorphisms to the product; I imagine a similar construction should be possible for double categories?

John Baez (Aug 06 2023 at 10:48):

@Morgan Rogers (he/him) - I guess your tensor product is different than Schappi's. You're categorifying the usual ⊗ of commutative monoids, e.g. the more familiar ⊗ of abelian groups. He's categorifying the usual ⊕ of commutative monoids.

In Theorem 5.2 he shows that given symmetric pseudomonoids M and M′ in a symmetric monoidal bicategory (C, ⊗), there is a natural way to make M ⊗ M′ into a symmetric pseudomonoid, of which the tensor product of 2-rigs is an example. Furthermore there is a 2-category of symmetric pseudomonoids in (C, ⊗), and this is cocartesian in the 2-categorical sense, with M ⊗ M′ being the 2-categorical coproduct of M and M′.

If we take (C, ⊗) = (Cat, $\times$), symmetric pseudomonoids in (Cat, $\times$) are symmetric monoidal categories, but Schappi's tensor product is just the usual product of categories M $\times$ M' made into a symmetric monoidal category. In Theorem 2.3 of Fong and Spivak's paper show that this is a biproduct in the 2-category SymmMonCat, so it deserves to be called ⊕. I expect this should work whenever (C, ⊗) is cartesian.

I would really like to see a good exploration of both the ⊗ and ⊕ for symmetric monoidal categories (or symmetric pseudomonoids), and how they interact.

John Baez (Aug 06 2023 at 11:05):

For example, ⊗ should distribute over ⊕ in a suitable sense, giving us a "rig 2-category" of symmetric monoidal categories.

James Deikun (Aug 06 2023 at 13:20):

I think they're actually categorifying basically the same thing and Schappi's is just more specialized.

John Baez (Aug 06 2023 at 13:35):

I don't know who "they" are, since we're talking about Morgan and Schappi and also Fong-Spivak. Fong-Spivak are more specialized than Schappi but they're getting a stronger result.

John Baez (Aug 06 2023 at 13:36):

But there really should be two quite different monoidal structures on SymmMonCat, with Fong-Spivak and Schappi studying the biproduct and Morgan studying another tensor product - just like how in AbGp we have $\oplus$ and $\otimes$.

Morgan Rogers (he/him) (Aug 06 2023 at 14:34):

John Baez said:

But there really should be two quite different monoidal structures on SymmMonCat, with Fong-Spivak and Schappi studying the biproduct and Morgan studying another tensor product - just like how in AbGp we have $\oplus$ and $\otimes$.

Exactly! Schappi's product requires no extra structure on the SMbiC, but generalizing the tensor product I'm describing to pseudomonoids in more general SMbiCs will require at least some colimits. I think the distributivity you're expecting will depend on how those colimits interact with the monoidal product... but this is a separate issue from how this tensor product might be generalised from monoidal categories to double categories!

James Deikun (Aug 06 2023 at 14:50):

Ah, sorry, you're mostly right here; I was looking at Theorem 5.1 in Schappi instead of 5.2. And playing around further it's clear the products are different. But I still can't see how the "Rogers monoidal product" could simultaneously be a categorification of $\otimes$ for commutative monoids, and also have the same set of objects as the product. Also generalizing this monoidal product to double categories would be tricky since it would first involve generalizing it away from symmetric monoidal categories to general ones.

Morgan Rogers (he/him) (Aug 06 2023 at 15:20):

James Deikun said:

But I still can't see how the "Rogers monoidal product" could simultaneously be a categorification of $\otimes$ for commutative monoids, and also have the same set of objects as the product.

In the same way that elements of the tensor product are presented by formal sums of pairs, but some such expressions are identified. Except that in this setting the natural notion of "identified" is free add isomorphisms.

As far as I can see, it should formally be possible to generalize to (non-symmetric) monoidal categories, since the symmetries (and their relation to the symmetries in the original monoidal categories) are also isomorphisms introduced after the fact... But I'm not confident about this since it's an operation I can't recall having seen an analogue of for ordinary monoids.

James Deikun (Aug 06 2023 at 15:43):

But there start out being way more formal sums of pairs than there are pairs; you can't get the tensor product just by identifying some pairs.

James Deikun (Aug 06 2023 at 15:46):

Although the idea I'm getting is less that your product isn't $\otimes$ and more that your original description of it wasn't accurate; would it be more accurate to describe it as taking the free symmetric monoidal category on $\mathbb{C} \times \mathbb{D}$ and throwing in some isomorphisms?

James Deikun (Aug 06 2023 at 16:00):

I think for algebraic theories having an internal hom requires the theory to be commutative, but I'm not sure the same can be said for a tensor product. https://groupprops.subwiki.org/wiki/Tensor_product_of_groups makes me lean in that direction though; you can't seem to get a single tensor product with just two groups as data that has the correct universal property.

John Baez (Aug 06 2023 at 16:05):

Ronnie Brown and others have studied some sort of tensor product of nonabelian groups, but I've never understood it.

Mike Shulman (Aug 06 2023 at 16:06):

"Having a tensor product" is a fairly vague phrase, but the obvious sort of tensor product that classifies "bi-homomorphisms" does require a commutative theory, for the same reasons as the obvious sort of internal-hom.

Mike Shulman (Aug 06 2023 at 16:10):

But symmetric monoidal categories are a "commutative 2-theory" in an appropriate sense (slightly weaker than the strict Cat-enriched notion). Tensor products for such 2-theories have been studied by various people, starting with Hyland and Power in "Pseudo-commutative monads and pseudo-closed 2-categories". Another useful reference is Bourke's "Skew structures in 2-category theory and homotopy theory".

Patrick Nicodemus (Aug 06 2023 at 18:19):

Mike Shulman said:

"Having a tensor product" is a fairly vague phrase, but the obvious sort of tensor product that classifies "bi-homomorphisms" does require a commutative theory, for the same reasons as the obvious sort of internal-hom.

Based on this comment, do you disagree with Morgan's hypothesis that there should be a notion of "tensor product of double categories" that classifies bi-double functors, i.e., functorial in vertical arrows on the left when the right arrow is held fixed? Because a double category generally cannot be "commutative" in the sense that a monoidal category can be symmetric.

Nathanael Arkor (Aug 06 2023 at 18:23):

A double category can be "symmetric" in the sense that there is an equivalence between the double category and the double category obtained by reversing its loose morphisms (generalising symmetry of a monoidal category when viewed as a one-object double category).

Patrick Nicodemus (Aug 06 2023 at 18:34):

Ok, that makes sense.

I think this concept is closest to my current situation. https://ncatlab.org/nlab/show/duoidal+category
But instead of two monoidal structures, I have two double category structures on the same category of arrows, and this distributive law holds wherever the compositions are defined. I wonder if anybody has written down a generalization of this concept from monoidal categories to double categories.

Patrick Nicodemus (Aug 06 2023 at 18:39):

I made a mistake earlier when I said this was "almost a double profunctor," I forgot some basic things about double profunctors this structure fails to have. It doesn't look much like a double profunctor at all.

Morgan Rogers (he/him) (Aug 06 2023 at 18:42):

James Deikun said:

Although the idea I'm getting is less that your product isn't $\otimes$ and more that your original description of it wasn't accurate; would it be more accurate to describe it as taking the free symmetric monoidal category on $\mathbb{C} \times \mathbb{D}$ and throwing in some isomorphisms?

Right, you would need to do that, since the isomorphism between $(X \oplus X',Y \oplus Y') \cong (X,Y) \oplus (X',Y')$ in the product is not compatible with the universal property we want to impose!

Morgan Rogers (he/him) (Aug 06 2023 at 18:45):

Mike Shulman said:

"Having a tensor product" is a fairly vague phrase, but the obvious sort of tensor product that classifies "bi-homomorphisms" does require a commutative theory, for the same reasons as the obvious sort of internal-hom.

Do you mean it needs to be commutative because we can externally exchange the arguments of a bi-hom and still have a bi-hom, and that transformation should be representable?

Mike Shulman (Aug 07 2023 at 07:25):

Patrick Nicodemus said:

Based on this comment, do you disagree with Morgan's hypothesis that there should be a notion of "tensor product of double categories" that classifies bi-double functors, i.e., functorial in vertical arrows on the left when the right arrow is held fixed?

I'm not sure exactly what that means. What would be the decategorified "tensor product of categories"?

Mike Shulman (Aug 07 2023 at 08:25):

Morgan Rogers (he/him) said:

Do you mean it needs to be commutative because we can externally exchange the arguments of a bi-hom and still have a bi-hom, and that transformation should be representable?

Unfortunately I don't remember offhand what exactly fails for a noncommutative theory. My guess would be that you might be able to construct a sort of "tensor product" that represents bihomomorphisms, but it won't be associative.

John Baez (Aug 07 2023 at 19:19):

I'd be interested if you can even do that! Like: is there a way to take any pair of groups $G$ and $H$ and get a group $G \star H$ and a function $i: G \times H \to G \star K$ that is a homomorphism in each argument, that is universal with this property? (For each group $K$ and each function $f : G \times H \to K$ that's a homomorphism in each argument, there exists a unique homomorphism $f': H \star G \to K$ such that $f = f' \circ i$.)

Mike Shulman (Aug 07 2023 at 19:29):

I think so. Let $U: \rm Grp \to Set$ be the underlying-set functor and $F$ its left adjoint. There are two functions $F(UG\times UH\times UH) \rightrightarrows F(UG \times UH)$, one of which multiplies in $H$ and then maps in, and the other of which maps in twice and then multiplies in the codomain. Similarly for $F(UG\times UG\times UH)$. The joint coequalizer of these two parallel pairs should be $G\star H$.

John Baez (Aug 07 2023 at 19:33):

Cool! It sounds reasonable when you say it: you take $F(UG \times UH)$ and mod out by just enough to force homomorphisms from this to any group $K$ be not arbitrary functions $UG \times UH \to K$ (which is what you'd get if you didn't mod out at all), but functions that are homomorphisms in each argument.

John Baez (Aug 07 2023 at 19:35):

I'm too lazy to check it, but I'm wondering if this is (a special case of) Ronnie Brown's tensor product of nonabelian groups. Unfortunately Brown considers some much more general kind of tensor product, where two groups each act on each other as automorphisms. But one could take both those actions to be trivial and see what his construction boils down to.

Mike Shulman (Aug 07 2023 at 19:36):

Interesting question. I don't know anything about Brown's tensor product. But I'm also either too busy or too lazy to check it.

John Baez (Aug 07 2023 at 19:39):

Maybe I'll take a look, because it's annoyed me for years - some guy says he knows how to take a tensor product of groups, and nobody else seems very interested...

Aaron David Fairbanks (Aug 08 2023 at 01:08):

I think this $G \star H$ is the tensor product of the abelianizations of $G$ and $H$. For all $g_1, g_2$ in $G$ and $h_1, h_2$ in $H$, we have
$(g_1g_2) \otimes (h_1h_2) = (g_1g_2 \otimes h_1)(g_1g_2 \otimes h_2) = (g_1 \otimes h_1)(g_2 \otimes h_1)(g_1 \otimes h_2)(g_2 \otimes h_2)$
and
$(g_1g_2) \otimes (h_1h_2) = (g_1 \otimes h_1h_2)(g_2 \otimes h_1h_2) = (g_1 \otimes h_1)(g_1 \otimes h_2)(g_2 \otimes h_1)(g_2 \otimes h_2)$.

So $(g_1 \otimes h_1)(g_2 \otimes h_1)(g_1 \otimes h_2)(g_2 \otimes h_2) = (g_1 \otimes h_1)(g_1 \otimes h_2)(g_2 \otimes h_1)(g_2 \otimes h_2)$.
Cancelling, we get $(g_2 \otimes h_1)(g_1 \otimes h_2) = (g_1 \otimes h_2)(g_2 \otimes h_1)$.

That means $G \star H$ is abelian. Multiplication in $G \star H$ gives elements of the form $g \otimes h$ with fixed $h$ the structure of a group under $G$, so all equations that hold in the abelianization of $G$ hold here, and similarly for $H$.

Aaron David Fairbanks (Aug 08 2023 at 01:12):

I don't think it's quite the same if you use monoids.

Mike Shulman (Aug 08 2023 at 02:51):

Nice argument! (Although I don't think elements of the form $g\otimes h$ with fixed $h$ are a group unless $h$ is the identity.)

In particular, this means there is no unit for this tensor product: if $G$ is nonabelian, there's nothing you can tensor it with to get itself back again, only its abelianization.

This is probably the best intuitive reason why we should expect to need a theory to be commutative in order to have a well-behaved tensor product.

Mike Shulman (Aug 08 2023 at 02:52):

That makes me wonder whether every theory has an "abelianization"? That is, for a theory $T$ is there a universal map into a commutative theory $T_{\rm ab}$, such that if $T$ is the theory of groups then $T_{\rm ab}$ is the theory of abelian groups?

Mike Shulman (Aug 08 2023 at 02:53):

If so, then maybe one can construct a "tensor product" of algebras for an arbitrary theory $T$ in a similar way, but it would turn out to actually be the tensor product of their "abelianizations", i.e. universal reflections into $T_{\rm ab}$-algebras.

Dan Marsden (Aug 08 2023 at 08:15):

If I recall correctly, the equations implied by commutativity (as a monad / theory) don't imply binary operations must be commutative. So the theory level commutativity of all derived operations doesn't imply the usual algebraic commutativity of a binary operation. So I'm not sure forcing a theory to be commutative will achieve the effect Mike suggests taking groups to Abelian groups, although there might be something specific to that case I'm missing.

Nathanael Arkor (Aug 08 2023 at 10:37):

I confused myself, so let me try again. Given an algebraic theory $T$, we can form a new algebraic theory $\mathrm{Comm}(T)$ by imposing the equations that force each pair of operations to commute with one another. This should provide a left adjoint to the inclusion of commutative algebraic theories into the category of algebraic theories. I think that the free commutative algebraic theory of the theory of groups should be the theory of abelian groups. There's something slightly misleading going on here, in that not every operation in $\mathrm{Comm}(T)$ must be commutative in the sense of being invariant under permutation of operands (we might call this "symmetry" rather than "commutativity" to avoid confusion). However, this turns out to be true in the case of groups, because imposing that the binary operation $\otimes$ commutes with itself forces $\otimes$ to be symmetric by the Eckmann–Hilton argument.

So while the theory of abelian groups is particularly well-behaved, I think taking the free commutative algebraic theory should still prove the appropriate generalisation of "abelianisation".

Nathanael Arkor (Aug 08 2023 at 10:44):

(In fact, I think it suffices for every operation in $T$ to have a unit in order that $\mathrm{Comm}(T)$ be symmetric, so this phenomenon is not unusual.)

John Baez (Aug 08 2023 at 10:50):

@Aaron David Fairbanks nicely blew a hole in my hopes for a 'tensor product of groups' with the universal property I described.

I believe a binary operation 'commutes with itself' in the sense of commutative monads if

$(g_1 g_2) (h_1 h_2) = (g_1 h_1) (g_2 h_2) \qquad \qquad \star$

If we have such an operation together with a constant that serves as an identity for this operation, then this operation must commute by the Eckmann-Hilton argument. No associativity is needed, so we could start with a unital magma, require that the multplication obey $\star$, and it would be forced to be commutative in the usual sense.

Aaron's argument starts by assuming that given magmas $G$ and $H$ there's a magma $G \otimes H$ such that

$g \otimes (h_1 h_2) = (g \otimes h_1) (g \otimes h_2)$

$(g_1 g_2) \otimes h = (g_1 \otimes h) (g_2 \otimes h)$

From this he gets

$(g_1g_2) \otimes (h_1h_2) = (g_1g_2 \otimes h_1)(g_1g_2 \otimes h_2)$
$= \big((g_1 \otimes h_1)(g_2 \otimes h_1)\big)\big((g_1 \otimes h_2)(g_2 \otimes h_2)\big)$

and

$(g_1g_2) \otimes (h_1h_2) = (g_1 \otimes h_1h_2)(g_2 \otimes h_1h_2)$
$= \big((g_1 \otimes h_1)(g_1 \otimes h_2)\big)\big((g_2 \otimes h_1)(g_2 \otimes h_2)\big)$

so

$\big((g_1 \otimes h_1)(g_2 \otimes h_1)\big)\big((g_1 \otimes h_2)(g_2 \otimes h_2)\big) =$
$\big((g_1 \otimes h_1)(g_1 \otimes h_2)\big)\big((g_2 \otimes h_1)(g_2 \otimes h_2)\big)$

To get further he assumes the operation is associative and has a unit and inverses - because he's working with groups! This lets him cancel stuff and get

$(g_2 \otimes h_1)(g_1 \otimes h_2) = (g_1 \otimes h_2)(g_2 \otimes h_1)$

showing that the operation in $G \otimes H$ is commutative.

John Baez (Aug 08 2023 at 10:56):

I'm not seeing how to carry out or modify Aaron's argument so it applies to monoids or more general magmas, in part because I'm completely exhausted by typing these equations and fixing all the typos in what I wrote. :sweat:

James Deikun (Aug 12 2023 at 15:45):

It seems like to even have a nice notion of what a "bihomomorphism" is, you need the monad to be strong, so you can draw diagrams like this in the base category:

$$\begin{CD} TG \times H @>{g \times H}>> G \times H @<{G \times h}<< G \times TH \\ @V{T\beta{} s}VV @VV{\beta}V @VV{T\beta{} t}V \\ TX @>>{x}> X @<<{x}< TX \end{CD}$$

James Deikun (Aug 12 2023 at 16:05):

And then to construct the tensor product $G \otimes H$ you want to coequalize the arrows from $F(TG \times H)$ to $F(G \times H)$ given as $F(g \times H)$ and $\mu Fs$ simultaneously with the arrows from $F(G \times TH)$ given as $F(G \times h)$ and $\mu Ft$, I believe.

James Deikun (Aug 12 2023 at 16:34):

Define a T-algebra $(G,g)$ as "Abelian" if for all X, Y, f, the following diagram commutes:

$$\begin{CD} T(TX \times Y) @<{t}<< TX \times TY @>{s}>> T(X \times TY) \\ @V{Ts}VV @. @VV{Tt}V \\ TT(X \times Y) @. \quad @. TT(X \times Y) \\ @V{\mu}VV @. @VV{\mu}V \\ T(X \times Y) @. \quad @. T(X \times Y) \\ @V{Tf}VV @. @VV{Tf}V \\ TG @>>{g}> G @<<{g}< TG \end{CD}$$

James Deikun (Aug 12 2023 at 17:21):

If the monad T is commutative, every T-algebra is Abelian as defined above. More generally, I think with a nice enough situation (something like "base category is cocomplete and small generated") you get an Abelianization for each individual algebra.

Define the Abelianization of a strong monad T as the universal commutative monad S with a strong lax monad morphism $\lambda : S \to T$. If the monad T itself has an Abelianization, I think every Abelian T-algebra defined as above inherits an algebra structure for the Abelianized monad via the universal factorization property of $\lambda$.

John Baez (Aug 13 2023 at 09:27):

Nice!!!

James Deikun (Aug 14 2023 at 20:51):

If you look at the tensor product $FX \otimes FY$ of two free algebras, and zoom in on the double coequalizer diagram, there is an arrow $T(\mu_X \times TY)$ from $F(TTX \times TY)$ to $F(TX \times TY)$. It has a section, $T(T\eta_X \times TY)$. If you chase naturality squares around, you get that the arrow $T(\eta_X \times TY)\mu_{X \times TY}Ts_{X,TY}$ is annihilated by the projection $\Pi$ from $F(TX \times TY)$ to $FX \otimes FY$. For similar reasons, so is $T(TX \times \eta_Y)\mu_{TX \times Y}Tt_{TX,Y}$. Obviously both of their composites are also annihilated.

If you further chase around naturality squares (so many naturality squares! :dizzy: ) you can put together an instance of the above Abelianism rectangle with $f = \eta_{TX \times TY} (\eta_X \times \eta_Y)$. Each leg starts out being an $\eta_{TX \times TY}$ followed by both of those loops in some order and you just hammer them into shape. The legs must be identified by postcomposition with $\Pi_{FX,FY}$.

James Deikun (Aug 14 2023 at 21:17):

Say you have some other $f : X \times Y \to T(G \times H)$. You can always express it as an arrow that factors through $\eta_{TX \times TY}(\eta_X \times \eta_Y)$. Using the naturality of $\Pi$ in the EM category, you can conclude that each component of $\Pi$ factors through Abelianization.

Patrick Nicodemus (Aug 16 2023 at 16:13):

Patrick Nicodemus said:

In this way $\mathbb{T}$ is almost a span from $\mathbb{C}^{\rm op}$ to $\mathbb{C}$ in double categories except that it's not a double category in a consistent way.

If anyone was curious, I realized the solution to this problem: the right setting is to work in a triple category, rather than a double category.

Essentially I had a triple category $\mathcal{E}$, and I was studying two embedded double categories inside it, trying to understand how they were related. My first double category $\mathcal{L}$ had

• category of objects $\mathcal{L}_O$ = horizontal 1-cells of $\mathcal{E}$, with horizontal 2-cells of $\mathcal{E}$ as horizontal morphisms
• category of arrows $\mathcal{L}_A$ = "basic" 2-cells of $\mathcal{E}$, with cubes between basic 2-cells of $\mathcal{E}$ as horizontal 2-cells in $\mathcal{L}$

Then the other double category $\mathcal{R}$ had

• category of objects $\mathcal{R}_O$ = vertical 1-cells of $\mathcal{E}$, with vertical 2-cells of $\mathcal{E}$ as horizontal morphisms
• category of arrows $\mathcal{R}_A$ = "basic" 2-cells of $\mathcal{E}$, with cubes between basic 2-cells of $\mathcal{E}$ as horizontal 2-cells in $\mathcal{R}$

In both cases you have the same category of arrows, but they are endowed with two different notions of vertical composition.
This new way of looking at it is a lot simpler, what a relief.

My next question is whether my weird "almost a double functor but not quite" is actually a triple functor.

Patrick Nicodemus (Aug 16 2023 at 16:13):

@James Deikun In case you are curious. :D

James Deikun (Aug 16 2023 at 16:28):

I was still curious in fact. Good thing you're not the type to give up when you see a triple category!

Patrick Nicodemus (Aug 16 2023 at 16:30):

I'm not happy about it.

Patrick Nicodemus (Aug 16 2023 at 16:30):

;)

Patrick Nicodemus (Aug 17 2023 at 00:59):

triple_categories_in_lifting_problems.pdf
ok I wrote up a quick sketch of this triple category. I would be very interested in knowing whether the constructions I define here are close to anything else in the literature. I tried to read the Grandis and Pare paper on intercategories briefly but didn't get too far into it. If anybody has experience with triple categories and wants to skim the note i would be happy

Mike Shulman (Aug 17 2023 at 02:34):

It looks like your "tensor product" of double categories can be described succinctly as (1) regard one double category as a horizontally discrete triple category, (2) regard the other one as a vertically discrete triple category, and (3) take their cartesian product as triple categories?

Patrick Nicodemus (Aug 17 2023 at 02:45):

Maybe, I'll think about that.

Patrick Nicodemus (Aug 17 2023 at 02:54):

Dare I ask if there is an agreed upon notion of a triple profunctor

Patrick Nicodemus (Aug 17 2023 at 02:56):

I guess you can always just use the definition from internal category theory.

Mike Shulman (Aug 17 2023 at 03:43):

Yeah. I don't think I've heard of triple profunctors before.

Patrick Nicodemus (Aug 17 2023 at 13:27):

The reason I ask is because based on your observation, I am speculating that my structure which was reminiscent (to me at least) a double profunctor J -|-> K might be in fact a triple profunctor from H(J) -> V(K), where H(J) is the vertically discrete triple category associated to J and V(K) the vertically discrete triple category associated to K.

I will look into this after work.

Patrick Nicodemus (Aug 17 2023 at 13:29):

In the paper above, I describe a triple category fibered over H(J)^op x V(K), and this is (I conjecture) some kind of tabulation or Grothendieck construction of a triple profunctor.

Patrick Nicodemus (Aug 17 2023 at 13:30):

In internal category theory profunctors are more or less defined to be their tabulations, I guess, so establishing this would not be too helpful.

Garner gives this notion of a coherent family of lifting functions between two diagrams indexed by double categories and I am trying to work out another way of describing this.

James Deikun (Aug 17 2023 at 14:47):

There's an interesting way to look at the lifting squares where you extend a square to an oriented 3-simplex -- one diagonal comes "for free" and another is the lifting. Then the condition of preserving the vertical composition becomes an "associativity" condition where you extend a pentagon to an oriented 4-simplex and make sure all the face maps give you the same faces that you already know (3 of the 5 faces come from lifts, 2 from the nerve of the base category).

James Deikun (Aug 18 2023 at 00:59):

Expanding on the above: take any number $n$ and any number $k \le n$. Create a simplex with $n$ ordered vertices. Label the first $k$ vertices with objects of $\mathbb{J}$ and the rest with objects of $\mathbb{K}$. Edges are typed: an edge between two $\mathbb{J}$-objects is a backwards vertical arrow of $\mathbb{J}$, an edge between two $\mathbb{K}$-objects is a forwards vertical arrow of $\mathbb{K}$, and an edge from a $\mathbb{J}$-object to a $\mathbb{K}$-object is an arrow of $\mathcal{C}$.

First, freely pick edges along the spine of the simplex. There will be either 0 or 1 arrows of $\mathcal{C}$ along the spine. If there is one, you also get to freely pick the edge from 0 to $n$. Then fill in the rest of the simplex. First fill in facets from the nerves of the vertical categories of $\mathbb{J}$ and $\mathbb{K}$. Then fill in facets which look like: a simplex from $\mathbb{J}$ and one from $\mathbb{K}$ glued to one from $\mathcal{C}$ along $U : \mathbb{J} \to \mathbb{S}\mathrm{q}(\mathcal{C})$ and $V : \mathbb{K} \to \mathbb{S}\mathrm{q}(\mathcal{C})$. Then fill in the facets which represent lifts: if they are tetrahedra fill in directly using the lifting structure, if they are higher-dimensional repeat this whole process on a smaller scale. If the lifting structure is well-behaved you should get the same results no matter which order you fill in facets.

James Deikun (Aug 18 2023 at 01:20):

Maps between simplices with the same $(n,k)$ consist of simplices built the same way, but with backwards horizontal arrows instead of objects of $\mathbb{J}$, forwards horizontal arrows instead of objects of $\mathbb{K}$; for edges, inverted squares of $\mathbb{J}$, normal squares of $\mathbb{K}$, and twisted squares of $\mathcal{C}$; and higher simplices the most important of which represent the naturality squares of the lifting structure.

Put together the simplices and their maps and you have a "nerve" for the lifting structure, displaying it as a kind of two-sided action of $U$ and $V$ as categories internal to somehow structured by $\mathrm{CAT}/\mathcal{C}$.

James Deikun (Aug 18 2023 at 14:52):

I'd like to think that there's some higher- or multi-categorical sense in which $\mathcal{\omega}$, the [[cosimplicial object]] of finite ordinals in CAT is "almost but not quite dense", and a double category over $\mathbb{S}\mathsf{q}(\mathcal{C})$ is an appropriate kind of "Segal structure" for the $\mathcal{\omega}$-nerve of $\mathcal{C}$ as an internal category. And then in the same sense, a lifting structure would be a "Segal structure" for the "twisted nerve" of $\mathrm{Hom}_{\mathcal{C}}$ as an internal "twisted bimodule".