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Stream: deprecated: algebra & CT

Topic: Why do bialgebras have monoidal module categories *really*

Chris Grossack (they/them) (Sep 14 2023 at 18:12):

Here's a cute question that I could probably answer myself if I had time to look up some facts that somebody here will just ~know~.

If $A$ is a $k$-algebra that's moreover a coalgebra in a compatible way then the module category $\text{Mod}(A)$ is monoidal, basically because if $V \otimes W$ is the usual tensor product of the underlying vector spaces, we can use the comultiplication $\Delta : A \to A \otimes A$ in order to act on $V \otimes W$.

Now, one of the big reasons to care about co-algebraic structures generally is that homming out of a co-thing gives you a thing. For instance, the circle $S^1$ is a cogroup (in the pointed homotopy category) so homming out of $S^1$ always gives a group (and indeed, "pointed homotopy classes of maps $S^1 \to X$" is called the fundamental group of $X$). Moreover, the $\mathsf{Mod}(A)$ functor from algebras to categories looks like homming out of $A$! Here's how to make that precise:

A $k$-algebra $A$ is the same data as a one-object $k$-linear category, and through this lens $\text{Mod}(A)$ is exactly the category of $k$-linear functors and natural transformations from $A$ to $\mathsf{Vect}$. So $\text{Mod}(A) = \underline{\text{Hom}}(A, \mathsf{Vect})$.

So let's look at $k\mathfrak{Lin}$, the 2-category of $k$-linear categories (read: categories enriched in $\mathsf{Vect}$). This has a tensor product, and if we identify $A$ with its one object category, then $A \otimes A$ as computed in $\mathsf{Vect}$ agrees with $A \otimes A$ as computed in $k\mathfrak{Lin}$. In particular, if $A$ is a bialgebra then it's a comonoid in $k\mathfrak{Lin}$.

Now, if all is right with the world, this should tell us that for any other object $C$ of $k\mathfrak{Lin}$ (like $\mathsf{Vect}$) $[A, C]$ should be a monoid in $\mathfrak{Cat}$. That is, a (strict) monoidal category! This also explains "automatically" why a cocommutative bialgebra should have a symmetric monoidal category worth of modules (read: a commutative monoid object in $\mathfrak{Cat}$). (Again, obviously there's a much more concrete way to see this using sweedler's notation, but let's ignore that for now).

I'm sure that there are subtleties that I'm missing here. For instance, if we want to remove the "strict" we should really be working with pseudomonoids. I'm sure that these work the same way, but I've never actually had the time to learn about them. Also, I think $\text{Mod}(-) = \underline{\text{Hom}}(-, \mathsf{Vect})$ should probably be valued in something a bit more restrictive than $\mathfrak{Cat}$. Maybe locally presentable (in the enriched sense) $k$-linear categories? Also also, this whole game looks like a $k$-linear version of functorial semantics. Has anyone written about that?

Thanks in advance! ^_^

John Baez (Sep 14 2023 at 19:36):

Chris Grossack (they/them) said:

If $A$ is a $k$-algebra that's moreover a coalgebra in a compatible way...

There are two things that phrase might mean: a [[bialgebra]] or a [[Frobenius algebra]]. But you mean a bialgebra since those are the algebras whose categories of modules become monoidal. Indeed, later you admit you mean a bialgebra.

John Baez (Sep 14 2023 at 19:37):

(A bialgebra is a coalgebra in the category of algebras. A Frobenius algebra is an algebra A that's a coalgebra in the category of A,A-bimodules.)

John Baez (Sep 14 2023 at 19:44):

I don't know who has written about the things you're talking about, but they seem pretty natural so someone has probably written about them at a high level of generality - like for enriched categories, not just Vect-enriched (= k-linear) categories. (I only mention this higher level of generality because it somehow seems more likely that people will have talked about your issues in this more general way!)

John Baez (Sep 14 2023 at 19:47):

For example suppose V is any cosmos. Then any monoid A in V is a one-object V-category and we get a V-category [A,V]. When V = Vect this is what you're calling Mod(A), so that's how you can think about it.

John Baez (Sep 14 2023 at 19:49):

When V is a bimonoid in V, [A,V] should be a monoidal V-category.

John Baez (Sep 14 2023 at 19:53):

Well, I could go on, but this goes a lot further than I could:

• Brian Day and Ross Street, Quantum categories, star autonomy, and quantum groupoids.

John Baez (Sep 14 2023 at 20:06):

Here's an interesting remark near the start:

A comonoidal category would have, instead of a tensor product, a tensor coproduct A → A × A and a counit with appropriately coherent constraints; this concept is not so interesting for ordinary categories but becomes more so for enriched categories. Comonoidal functors would go between comonoidal categories.

Chris Grossack (they/them) (Sep 14 2023 at 20:07):

John Baez said:

When V is a bimonoid in V, [A,V] should be a monoidal V-category.

Yeah, this is exactly the kind of result I'm looking for! I was sure someone had thought about it, and it's believable that people would be quick to do this in the general enriched setting. Especially through the lens of some $\mathcal{V}$-enriched lawvere duality (or more likely, gabriel-ulmer duality), a la Kelly and Power.

Chris Grossack (they/them) (Sep 14 2023 at 20:08):

John Baez said:

Well, I could go on, but this goes a lot further than I could:

• Brian Day and Ross Street, Quantum categories, star autonomy, and quantum groupoids.

Thanks! I'll give this a read later today ^_^

John Baez (Sep 14 2023 at 20:09):

It's probably too much.... and also not enough of what you were asking for.

Chris Grossack (they/them) (Sep 14 2023 at 20:13):

Isn't it always, haha. If push comes to shove I'll work through the details myself when I have a bit of free time, and probably write a blog post on it. Hopefully someone is able to just tell me what's going on though, haha. That would be much faster :P

John Baez (Sep 14 2023 at 20:15):

Here's one thing I just wanted to add, that wasn't in your story. If V is a cosmos and M is a monoidal V-category then [M,V] becomes a monoidal V-category using [[Day convolution]]. (In fact the nLab explains it at this level of generality.)

John Baez (Sep 14 2023 at 20:16):

One might naively have thought M needed some 'comonoidal' structure to make [M,V] monoidal.

Chris Grossack (they/them) (Sep 14 2023 at 20:17):

John Baez said:

Here's one thing I just wanted to add, that wasn't in your story. If V is a cosmos and M is a monoidal V-category then [M,V] becomes a monoidal V-category using [[Day convolution]].

I don't think I knew that Day Convolution worked over any cosmos V. I'd only seen it in the case of $V = \mathsf{Set}$! It's literally the second sentence on the nlab page, lol, so I guess I didn't read that very closely

Simon Willerton (Sep 14 2023 at 20:18):

I think that if V is a cosmos, then a bimonoid in V can be thought of a one-object comonoidal V-category. (You don't normally hear about comonoidal categories because when V=Set the monoidal structure is cartesian so every Set-category is comonoidal.)

John Baez (Sep 14 2023 at 20:18):

(And I bet they're probably all comonoidal in just one way, which makes it sort of 'boring'.)

Chris Grossack (they/them) (Sep 14 2023 at 20:18):

John Baez said:

One might naively have thought M needed some 'comonoidal' structure to make [M,V] monoidal.

That's a really good point, actually... I did (naively :P) expect $M$ to have some comonoidal structure... But I think I can recover because comonoidal structure on $M$ should be "the same thing" as a monoidal structure on every $[M,X]$, naturally in $X$!

Simon Willerton (Sep 14 2023 at 20:20):

Every comonoidal V-category M is an example of a promonoidal V-category and promonoidal structures on V-categories correspond precisely to Day convolution products on [M, V], and that's the category of modules of the bimonoid M.

John Baez (Sep 14 2023 at 20:21):

Chris Grossack (they/them) said:

John Baez said:

One might naively have thought M needed some 'comonoidal' structure to make [M,V] monoidal.

That's a really good point, actually... I did (naively :P) expect $M$ to have some comonoidal structure... But I think I can recover because comonoidal structure on $M$ should be "the same thing" as a monoidal structure on every $[M,X]$, naturally in $X$!

Umm, but see what Simon said in the case of V = Set. Every category C is comonoidal, but that doesn't make [C,Set] monoidal. (I don't claim to truly understand what's going on here!)

Chris Grossack (they/them) (Sep 14 2023 at 20:21):

Simon Willerton said:

I think that if V is a cosmos, then a bimonoid in V can be thought of a one-object comonoidal V-category. (You don't normally hear about comonoidal categories because when V=Set the monoidal structure is cartesian so every Set-category is comonoidal.)

I'm fairly sure this is true, basically because Kelly's tensor product of V-categories agrees with the usual tensor product in V, provided you're looking at the one-object V-categories (read: V-monoids).

Chris Grossack (they/them) (Sep 14 2023 at 20:23):

Simon Willerton said:

Every comonoidal V-category M is an example of a promonoidal V-category and promonoidal structures on V-categories correspond precisely to Day convolution products on [M, V], and that's the category of modules of the bimonoid M.

I'm going to have to think really hard about this, haha. I don't know much about "promonoidal structures" at all. But I'm definitely happy with [M,V] being modules for the bimonoid M, haha.

Simon Willerton (Sep 14 2023 at 20:24):

(The "I think" means I've been out to dinner and can't be bothered to check the details, but I'm pretty sure.)

John Baez (Sep 14 2023 at 20:24):

Simon Willerton said:

Every comonoidal V-category M is an example of a promonoidal V-category and promonoidal structures on V-categories correspond precisely to Day convolution products on [M, V]....

How does that mesh with what you said earlier: for V = Set every category C is comonoidal? What monoidal structure does that put on [C,Set]? Oh - maybe something 'dull' like products or coproducts?

Simon Willerton (Sep 14 2023 at 20:25):

Look in Day's thesis for the promonoidal stuff, in particular the example of comonoidal categories, or else the Day-Street reference John gave.

Simon Willerton (Sep 14 2023 at 20:25):

For V=Set, you have the pointwise product of preseheaves.

John Baez (Sep 14 2023 at 20:25):

Okay, 'twas staring me in the face.

John Baez (Sep 14 2023 at 20:35):

Chris Grossack (they/them) said:

Simon Willerton said:

Every comonoidal V-category M is an example of a promonoidal V-category and promonoidal structures on V-categories correspond precisely to Day convolution products on [M, V], and that's the category of modules of the bimonoid M.

I'm going to have to think really hard about this, haha. I don't know much about "promonoidal structures" at all.

It definitely sounds terrifying, but it's not so terrible if you understand [[profunctors]] and are willing to accept that you can work with profunctors very similarly in the V-enriched case. A V-enriched profunctor from a V-category $C$ to a V-category $D$ is a V-enriched profunctor from $C$ to the presheaf category $[D^{\mathrm{op}},V]$.

John Baez (Sep 14 2023 at 20:36):

There's a 2-category VProf of V-enriched categories, V-enriched profunctors and V-enriched natural transformations.

John Baez (Sep 14 2023 at 20:37):

A "promonoidal" category is like a monoidal category, but where the tensor product etc. are profunctors instead of functors!

John Baez (Sep 14 2023 at 20:37):

If you want to scare people, just say "a promonoidal V-category is a pseudomonoid in VProf".

John Baez (Sep 14 2023 at 20:39):

But V-enriched profunctors from $C$ to $D$ also correspond to cocontinuous V-functors from $[C^{\rm op},V]$ to $[D^{\rm op},V]$. You may know this in the case V = Set.

John Baez (Sep 14 2023 at 20:39):

So, a promonoidal structure on $C$ is the same as a (V-enriched) monoidal structure on $[C^{\rm op},V]$ that's cocontinuous in each argument.

Simon Willerton (Sep 17 2023 at 19:33):

PS. This perhaps is clear to many, but I will mention it anyway. A commutative algebra (or "monoid"), A, in V can be considered as one-object monoidal V-category; a monoidal category is also promonoidal, thus the Day product then gives a (biclosed) monoidal structure to A's representation category [A, V]. This should be familiar, but noticeably different to the monoidal product on the representation category of a bialgebra.

Simon Willerton (Sep 17 2023 at 19:33):

In the world of categories, bialgebras are kind of dual to commutative algebras.

John Baez (Sep 18 2023 at 08:39):

I occasionally get disturbed by the fact that both commutative algebras and bialgebras get a monoidal structure on their representation categories... especially when I'm working with a commutative bialgebra.

John Baez (Sep 18 2023 at 10:28):

For example take $\mathbb{C}[x]$, the free algebra over $\mathbb{C}$ on one generator.

John Baez (Sep 18 2023 at 10:29):

A module of this is just a (complex) vector space $V$ with a linear map $T: V \to V$.

John Baez (Sep 18 2023 at 10:30):

Since $\mathbb{C}[x]$ is commutative, its category of modules gets a tensor product.

John Baez (Sep 18 2023 at 10:30):

But I think we can make $\mathbb{C}[x]$ into a bialgebra in (at least) two ways.

John Baez (Sep 18 2023 at 10:31):

The comultiplication is always determined by what it does to $x$, and one choice is

$x \mapsto x \otimes x$

John Baez (Sep 18 2023 at 10:32):

This makes $\mathbb{C}[x]$ into a bialgebra which is the monoid algebra of the monoid $(\mathbb{N}, +, 0)$.

John Baez (Sep 18 2023 at 10:33):

So here the modules of $\mathbb{C}[x]$ get viewed as representations of $\mathbb{N}$ and we're getting a way to tensor them which is the usual way of tensoring monoid representations.

John Baez (Sep 18 2023 at 10:34):

This is different from the previous tensor product, since now the unit for the tensor product is $\mathbb{C}$, while before it was $\mathbb{C}[x]$.

John Baez (Sep 18 2023 at 10:35):

But I think there's a second bialgebra structure on $\mathbb{C}[x]$, determined by

$x \mapsto x \otimes 1 + 1 \otimes x$.

John Baez (Sep 18 2023 at 10:35):

This makes $\mathbb{C}[x]$ into a Hopf algebra which is the universal enveloping algebra of the 1-dimensional abelian Lie algebra $\mathbb{C}$.

John Baez (Sep 18 2023 at 10:37):

So here the modules of $\mathbb{C}[x]$ get viewed as representations of the Lie algebra $\mathbb{C}$ and we're getting a way to tensor them which is the usual way of tensoring Lie algebra representations!

John Baez (Sep 18 2023 at 10:37):

Unless I'm confused this is different from both the previous tensor products! (Is it?)

John Baez (Sep 18 2023 at 10:37):

But as you can see, this thicket of monoidal structures gets confusing.

John Baez (Sep 18 2023 at 10:38):

So I could have made some mistakes....