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Stream: theory: category theory

Topic: Non-symmetric monoidal monads

Naïm Favier (Mar 08 2024 at 14:40):

I asked a question on MSE about the existence of non-symmetric monoidal monads, in case someone here might know.

Sam Staton (Mar 09 2024 at 08:48):

Interesting question: is there a monoidal monad on an smc that is not symmetric monoidal?

One approach is to start from a non-symmetric monoidal functor between smc's $F:C \to D$. Let me write $\hat C=[C^{op},\mathbf{Set}]$. Then we have smc's $\hat C$ and $\hat D$ with Day convolution, and by Kan extension a monoidal functor $L:\hat C \to \hat D$ between smc's , which moreover has a right adjoint $R:\hat D\to \hat C$ and hence a monad $RL$ on $\hat C$. I think this will give a monoidal structure for $RL$ but not necessarily a symmetric monoidal one. Did you try this? I think it is what @Jonas Frey and @Richard Williamson were discussing on nforum a few years ago.

To be concrete, I just tried this for the non-symmetric monoidal functor $\mathbb{N}\to\mathbb{B}$ where $\mathbb{N}$ is the discrete category of natural numbers and $\mathbb{B}$ is the category of natural numbers and bijections; these are both smc's. The functor seemed to give an example of a monoidal monad of the kind you were looking for. i.e. a monoidal monad on $[\mathbb{N},\mathbf{Set}]$ that is not symmetric monoidal, wrt the Day convolution symmetric monoidal structure. I think I can see the two different strengths coming out of the construction.
(Of course, this is subtle and I may have made a mistake.)

Tobias Fritz (Mar 09 2024 at 09:53):

Here's a different approach that is perhaps worth considering. If $M$ is a commutative monoid in any symmetric monoidal category, then $M \otimes -$ is a monoidal monad in a canonical manner. Now choose a different symmetric braiding on the underlying monoidal category. Then generically we should expect $M \otimes -$ to no longer be symmetric, right? This happens as soon as the monoid $M$ is not commutative with respect to the new symmetric braiding.

Tobias Fritz (Mar 09 2024 at 09:58):

For a concrete example, consider the monoidal category of [[super-vector space]], which has two canonical symmetric braidings. Any $\mathbb{Z}_2$-graded commutative algebra should be a commutative monoid in this category with respect to the traivial braiding (see nLab page), but it will not be commutative with respect to the super-braiding (aka "super-commutative") unless the odd part is trivial.

Tobias Fritz (Mar 09 2024 at 09:58):

Does this work? Like Sam, I haven't checked all the details :sweat_smile:

Naïm Favier (Mar 09 2024 at 09:58):

I thought of that, but now you have to find a monoidal category with two different symmetric braidings and a commutative monoid on one of them, and I don't have a solid enough math background to think of any examples. I didn't even know the former was possible until like yesterday.

Another thing i thought about is to consider the free symmetric monoidal category on a monoidal category that has a monoidal monad; then there should be no reason to expect it to be symmetric with respect to that, but i have no idea how free symmetric monoidal categories on monoidal categories work.

@Sam Staton 's idea seems promising, i'll play around with it later.

Naïm Favier (Mar 09 2024 at 09:59):

Thanks for the example

Dylan McDermott (Mar 09 2024 at 12:59):

There is also an example on page 133 of What makes a strong monad? (though it isn't explicitly written out as a monoidal monad there). If $M$ is a commutative monoid in $\mathbf{Set}$, then we can consider the cartesian monoidal category $\mathsf{Act}M$ of sets $Y$ equipped with monoid action $(y, m) \mapsto y \ast m$. Then $M$ is a monoid in $\mathsf{Act} M$, so $TX = X \times M$ is a monad on $\mathsf{Act}M$. You can make this into a non-symmetric monoidal monad in which the map $TX \times TY \to T(X \times Y)$ sends $((x, m), (y, m'))$ to $(x, y \ast m, m \cdot m')$. This might be an instance of what @Tobias Fritz suggested.

Naïm Favier (Mar 09 2024 at 16:16):

Thanks, @Dylan McDermott , that's exactly the sort of thing i was looking for. I'll add that as an answer to my MSE question.

John Baez (Mar 10 2024 at 01:31):

Tobias Fritz said:

Here's a different approach that is perhaps worth considering. If $M$ is a commutative monoid in any symmetric monoidal category, then $M \otimes -$ is a monoidal monad in a canonical manner. Now choose a different symmetric braiding on the underlying monoidal category. Then generically we should expect $M \otimes -$ to no longer be symmetric, right? This happens as soon as the monoid $M$ is not commutative with respect to the new symmetric braiding.

For some reason - not a good reason - I feel like using the category $\mathsf{Vect}[\mathbb{Z}]$ of $\mathbb{Z}$-graded vector spaces. Just like the category $\mathbb{Z}_2$-graded vector spaces this has a standard monoidal structure which extends to two different symmetric monoidal structures -one using the "boring" symmetry (which is not "trivial" in the strongest possible sense, but still boring) and the other using the "super" symmetry.

The exterior algebra $\Lambda V$ on a vector space is a monoid in $\mathbb{Z}$-graded vector spaces that's commutative only with respect to the "super" symmetry (unless $V$ is at most 1-dimensional).

So concretely, are you suggesting that $\Lambda V \otimes -$ is a monoidal monad on $\mathsf{Vect}[\mathbb{Z}]$ that's not a symmetric monoidal monad with respect to the "boring" symmetry on $\mathsf{Vect}[\mathbb{Z}]$?

John Baez (Mar 10 2024 at 01:34):

It's a monad because tensoring with a monoid object is a monad. That much I know.

Then it seems you're claiming that it's a symmetric monoidal monad if we use the "super" symmetry on $\mathsf{Vect}[\mathbb{Z}]$.

Is there some general result that implies this, namely: tensoring with a commutative monoid object in a symmetric monoidal category is a symmetric monoidal monad?

John Baez (Mar 10 2024 at 01:34):

(I've never thought much about monoidal or symmetric monoidal monads.)

John Baez (Mar 10 2024 at 01:36):

And then I guess you're also claiming that every symmetric monoidal monad is a commutative monad... and that $\Lambda V \otimes -$ is not a symmetric monoidal monad on $\mathsf{Vect}[\mathbb{Z}]$ with its "boring" symmetry.

John Baez (Mar 10 2024 at 01:38):

I suppose if I wanted to master the coherence laws for monoidal and symmetric monoidal monads, checking these things would be very good practice! Unfortunately I'm a lazy bum.

Tobias Fritz (Mar 10 2024 at 05:39):

John Baez said:

So concretely, are you suggesting that $\Lambda V \otimes -$ is a monoidal monad on $\mathsf{Vect}[\mathbb{Z}]$ that's not a symmetric monoidal monad with respect to the "boring" symmetry on $\mathsf{Vect}[\mathbb{Z}]$?

Yes, that's a nice example!

Tobias Fritz (Mar 10 2024 at 05:47):

John Baez said:

Is there some general result that implies this, namely: tensoring with a commutative monoid object in a symmetric monoidal category is a symmetric monoidal monad?

Right! If we write $M$ for the monoid and $T := M \otimes -$ for the monad, then we first of all need a natural transformation $\nabla : TX \otimes TY \to T(X \otimes Y)$ to define the lax monoidal structure maps. This amounts to $M \otimes X \otimes M \otimes Y \to M \otimes X \otimes Y$, which we can define as swapping the middle two factors first and then using the multiplication in $M$.

Tobias Fritz (Mar 10 2024 at 05:47):

Now why would we want the monoid to be commutative? This has to do with the required compatibility of $\nabla$ with the monad multiplication:

Tobias Fritz (Mar 10 2024 at 05:48):

image.png
(I'm lazy too and copied this from here, where we were in the cartesian monoidal setting, so there should be $\otimes$ in place of $\times$.)

Tobias Fritz (Mar 10 2024 at 05:51):

This diagram amounts to starting with $M \otimes M \otimes M \otimes M$ and multiplying down to $M$ in the lower left composite, while the upper right composite in addition involves two swaps. So I think that this commutes if and only if $M$ is commutative.

John Baez (Mar 10 2024 at 05:51):

That's cool! I'll have to think about that.

This is what I was writing while you were writing that:

A nice thing is that when we've got a monad on a symmetric monoidal category $\mathsf{M}$ of the form

$m \otimes -$

for some commutative monoid object $m \in \mathsf{M}$, this monad arises from an adjunction between $\mathsf{M}$ and the category of $m$-actions in $\mathsf{M}$ - or you might call them "$m$-modules". The left adjoint sends any object $x \in \mathsf{M}$ to the free $m$-module on that, which is $m \otimes x$.

So, our friend

$\Lambda V \otimes -$

on $\mathsf{Vect}[\mathbb{Z}]$ arises from an adjunction between the category of graded vector spaces and the category of graded $\Lambda V$-modules... where $\Lambda V \otimes -$ maps any graded vector space to the free graded $\Lambda V$-module on that graded vector space.

I think the category of $\Lambda V$-modules is the Eilenberg-Moore category of this monad.

Now, if this monad is commutative, what does that imply about its Eilenberg-Moore category? Does it get a symmetric monoidal structure?

Tobias Fritz (Mar 10 2024 at 05:56):

John Baez said:

And then I guess you're also claiming that every symmetric monoidal monad is a commutative monad... and that $\Lambda V \otimes -$ is not a symmetric monoidal monad on $\mathsf{Vect}[\mathbb{Z}]$ with its "boring" symmetry.

Yes, though since I didn't check all the coherences either, I'm a bit uncomfortable with "claiming" that and prefer your earlier phrasing of "suggesting" :laughing:

John Baez (Mar 10 2024 at 05:59):

Whoops, I'm starting to copy Jim Dolan, who likes to jokingly claim that I'm "claiming" any idea he thinks of that might follow from things I said or guessed or thought about out loud.

Tobias Fritz (Mar 10 2024 at 05:59):

As a caveat, it's worth noting that there are two slightly different notions of "commutative monad", as per Remark 2.1 on the nLab page: "in the setting where $V$ is symmetric monoidal, we will assume that the left and right strengths are related by the symmetry in the obvious way". So this thread is all about how this additional property can fail.

Tobias Fritz (Mar 10 2024 at 06:04):

John Baez said:

So, our friend

$\Lambda V \otimes -$

on $\mathsf{Vect}[\mathbb{Z}]$ arises from an adjunction between the category of graded vector spaces and the category of graded $\Lambda V$-modules... where $\Lambda V \otimes -$ maps any graded vector space to the free graded $\Lambda V$-module on that graded vector space.

For context, are there important examples of $\Lambda V$-modules? I like exterior algebras, but I'm not sure I've ever come across any of their modules in nature. Is this related to your work on Schur functors and plethysm?

Tobias Fritz (Mar 10 2024 at 06:08):

Now, if this monad is commutative, what does that imply about its Eilenberg-Moore category? Does it get a symmetric monoidal structure?

I'm personally more familiar with Kleisli categories, since that is what's most relevant for probability. And there indeed yes, a commutative monad is exactly what we need in order to turn the Kleisli category into a symmetric monoidal category. But the funny thing is that the commutativity is not just relevant for the symmetry there, but already for the Kleisli category to be monoidal at all. (This somehow reminds me of the periodic table and stabilization hypothesis, but that's just a feeling I get.)

If the monad is merely monoidal, then the Kleisli category is merely premonoidal, meaning that the interchange law may not hold. Computer scientists are interested in this because side effects generally don't satisfy the interchange law. I guess that this applies to actions of agents on an environment as well: this "side effect" generally depends on the order of the agent's actions, even if the agents are disjoint.

Tobias Fritz (Mar 10 2024 at 06:11):

But also the Eilenberg-Moore category is symmetric monoidal if you have suitable coequalizers -- this generalizes the construction of the tensor product of modules over a commutative ring as a coequalizer of two maps, one with the ring acting on the first module and one with the ring acting on the second module. This also goes back to Anders Kock, and Chris Heunen explained it here on MO.

John Baez (Mar 10 2024 at 06:12):

I haven't really thought about exterior algebra modules too much. We can compare symmetric algebra modules. The symmetric algebra $S V^*$ on the dual of a vector space $V$ is the algebra of polynomial functions on $V$. So a finitely generated projective $S V^*$-module is a kind of "algebraic vector bundle" over $V$.

In analogy, if we like supersymmetry, we say $\Lambda V^*$ is the algebra of polynomial functions on the odd, or fermionic vector space $V$.

So then a finitely generated projective $\Lambda V^*$-module is a kind of algebraic vector bundle on this odd vector space!

John Baez (Mar 10 2024 at 06:19):

I haven't thought about $\Lambda V$-modules when thinking about Schur functors and plethym. But I probably should. The free commutative algebra on one generator, namely $k[x]$ where $k$ is our favorite field, is an example of a "plethory": besides being a commutative algebra it has an extra operation, namely that we can compose polynomials in one variable.

The free exterior algebra on one generator, namely $k[x]/\langle x^2 \rangle$, is the free supercommutative algebra on one odd generator. So they're like the fermionic version of polynomials in one generator - "functions on the odd line". Can we compose these guys? Does this make $k[x]/\langle x^2 \rangle$ into a plethory? I don't know.

Tobias Fritz (Mar 10 2024 at 06:26):

Well I guess we can compose $x \mapsto a x + b$ and $x \mapsto c x + d$ to $x \mapsto c (a x + b) + d = acx + (cb + d)$? Meaning that linear functions are already closed under composition (unlike quadratic or higher). But then I don't know what further compatibility conditions are required of a plethory.

John Baez (Mar 10 2024 at 06:29):

They're sort of long to state, but I'm thinking everything will work because they work for polynomials and here we can do the same things with polynomials of the form $a x + b$ and then discard terms involving $x^2$ or higher powers.

John Baez (Mar 10 2024 at 06:30):

Add them, multiply them, compose them, and other stuff I'm not mentioning!

Naïm Favier (Mar 10 2024 at 09:11):

John Baez said:

And then I guess you're also claiming that every symmetric monoidal monad is a commutative monad...

I've checked all the things in Agda here. The one-line summary is that commutative monads are exactly monoidal monads, and on top of that symmetric commutative monads are exactly symmetric monoidal monads (where by symmetric commutative monad i mean that the strengths are related by the braiding).

Naïm Favier (Mar 10 2024 at 09:16):

There seems to be an implicit convention of saying "commutative monad" to mean "symmetric commutative monad", which hopefully this thread (and this older one) make clear is harmful :upside_down:

Naïm Favier (Mar 10 2024 at 09:19):

BTW, the linked 1Lab page has an explanation of "monoidal implies commutative" in the style of an Eckmann-Hiltonish argument; have you seen this before?

Tobias Fritz (Mar 10 2024 at 09:26):

Naïm Favier said:

There seems to be an implicit convention of saying "commutative monad" to mean "symmetric commutative monad", which hopefully this thread (and this older one) make clear is harmful :upside_down:

Do you know of any significance for or applications of non-symmetric commutative monads? Because I work with symmetric commutative monads a lot, and having to add the "symmetric" all the time would be quite cumbersome, so I'm reluctant to do it. And as long as there aren't many non-symmetric examples of interest, I don't see the harm in keeping the terminology as it's mostly used.

Tobias Fritz (Mar 10 2024 at 09:27):

There are a lot of situations like this in category theory. For example, one could object that you should really be saying "lax monoidal monad" instead of "monoidal monad", since for many people monoidal functors are strong monoidal by default.

Naïm Favier (Mar 10 2024 at 09:31):

Fair enough, but at least the nuance should be explained when first introducing the concept, and one should not oppose "commutative" and "monoidal", but rather "symmetric" and "non-symmetric"

Sam Staton (Mar 12 2024 at 20:56):

Fascinating discussion and examples. I went over the reply I wrote and it seems to check out too.
So: Species ($[\mathrm{Bij},\mathbf{Set}]$) are monadic over $[\mathbb{N},\mathbf{Set}]$, and this monad on $[\mathbb{N},\mathbf{Set}]$ is monoidal, but not symmetric monoidal, even though $[\mathbb{N},\mathbf{Set}]$ can be thought of as symmetric monoidal. (Here I am considering the Day tensor i.e. multiplication of species.)

Sam Staton (Mar 12 2024 at 20:56):

More generally, it seems a PRO can be regarded as a colimit-preserving monoidal monad on $[\mathbb{N},\mathbf{Set}]$, which will almost never be a symmetric monoidal monad unless it is a commutative monoidal category.

Sam Staton (Mar 12 2024 at 20:56):

I suspect some versions of cubical sets are monadic over $[\mathrm{Bij},\mathbf{Set}]$ with a monoidal but not symmetric monoidal monad.

Sam Staton (Mar 12 2024 at 20:57):

Actually @Naïm Favier at MSE you said they were "failing to answer" at nforum but are you sure? Arguably they said things like what I said, just in a chatty way.

Naïm Favier (Mar 13 2024 at 09:18):

Sam Staton said:

Fascinating discussion and examples. I went over the reply I wrote and it seems to check out too.
So: Species ($[\mathrm{Bij},\mathbf{Set}]$) are monadic over $[\mathbb{N},\mathbf{Set}]$, and this monad on $[\mathbb{N},\mathbf{Set}]$ is monoidal, but not symmetric monoidal, even though $[\mathbb{N},\mathbf{Set}]$ can be thought of as symmetric monoidal. (Here I am considering the Day tensor i.e. multiplication of species.)

I see, so in this case the strengths correspond to extending an element of $S_k$ to an element of $S_n$ by either "padding" on the left or on the right. These are distinct operations, yet they commute in the sense that right-padding $a : S_k$ and left-padding $b : S_l$ give commuting elements of $S_{k+l}$. Interesting!

Sam Staton (Mar 14 2024 at 17:00):

Thanks for checking, yes that's right. I'm glad you raised the question --- for some years we have been looking at algebraic theories enriched in presheaf categories (most recently here) and I had been meaning to check what is strange with the symmetry of the monoidal structure on $\mathbf{Set}^{\mathbb{N}}$ since I think Cristina also brought it up.