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Stream: theory: category theory

Topic: 2-categories of symmetric strict monoidal categories

Chad Nester (Mar 07 2022 at 14:00):

Let $\mathsf{StrSMC}$ be the (strict) 2-category of symmetric strict monoidal categories, symmetric strict monoidal functors, and monoidal natural transformations. Let $\mathsf{SMC}$ be the version of this where the 1-cells are instead symmetric strong monoidal functors.

Is $\mathsf{StrSMC}$ equivalent to $\mathsf{SMC}$?

If we take lax monoidal functors then I know we get an adjunction (strictification theorem). We certainly get that here as well, but I'm wondering if asking for the 1-cells to be strong monoidal is enough to make it an equivalence.

Some evidence against is that $\mathsf{SMC}(\mathbb{X},\mathbb{Y})$ is symmetric strict monoidal with $F \otimes G$ defined as in $(F \otimes G)(X) = FX \otimes GX$, but that $\mathsf{StrSMC}(\mathbb{X},\mathbb{Y})$ is not, since this would mean that $FA \otimes GA \otimes FB \otimes GB = (F \otimes G)(A \otimes B) = FA \otimes FB \otimes GA \otimes GB$, which isn't true in general. Is this enough to show that they're not equivalent?

Chad Nester (Mar 07 2022 at 14:03):

For context, I've been working through this paper on tensor product structure for symmetric monoidal categories. While it deals with the lax monoidal case of this situation, much of it also applies to the strong monoidal case. In particular we have $\mathsf{StrSMC}(\mathbb{X} \otimes \mathbb{Y}, \mathbb{Z}) \cong \mathsf{SMC}(\mathbb{X},\mathsf{SMC}(\mathbb{Y},\mathbb{Z}))$ -- an isomorphism!

Chad Nester (Mar 07 2022 at 14:05):

I was kind of expecting an equivalence $\mathsf{SMC}(\mathbb{X} \otimes \mathbb{Y},\mathbb{Z}) \simeq \mathsf{SMC}(\mathbb{X},\mathsf{SMC}(\mathbb{Y},\mathbb{Z}))$, and I don't really understand why this isn't true (if indeed it is not).

Mike Shulman (Mar 07 2022 at 15:33):

I don't have a counterexample to hand, but in general the answer to this sort of thing is no.

Chad Nester (Mar 07 2022 at 15:34):

Thanks! This is still very helpful.

Martti Karvonen (Mar 07 2022 at 17:26):

Cartesian products give (2-categorical) biproducts for $\mathsf{SMC}$, see e.g. the appendix of this. As cartesian products are products also in $\mathsf{StrSMC}$, to prove that the two 2-categories not equivalent it should be sufficient to show that the product is not a biproduct in $\mathsf{StrSMC}$. First of all, the injection maps are defined by their projections so we can assume they behave in $\mathsf{StrSMC}$ as they do in $\mathsf{SMC}$, i.e. the left injection sends an object $A$ to $(A,I)$ and similarly for the right injection. One can then cook up situations where there is no symmetrict strict monoidal map from $C\times D$ that makes the required diagrams commute up to iso, even when the starting maps from $C$ and $D$ are strict (for instance, the "codiagonal" corresponds to the monoidal product, which is rarely isomorphic to a symmetric strict monoidal functor). Thus the product fails to be a coproduct in $\mathsf{StrSMC}$.

John Baez (Mar 07 2022 at 17:44):

Nice argument! I "knew" these 2-categories couldn't be equivalent, but it would have taken me work to find a proof.

So you're saying that for a strict symmetric monoidal category $C$, the monoidal product $\otimes : C \times C \to C$ is usually not monoidally naturally isomorphic to a symmetric strict monoidal functor.

Let me think about why. For it to actually be a strict monoidal functor we'd need

$(c \otimes d) \otimes (c' \otimes d') = (c \otimes c') \otimes (d \otimes d')$

and various related things. So these are the things we want to break.

Martti Karvonen (Mar 07 2022 at 18:52):

I guess there's still some work left to do: the monoidal product is rarely strict itself, but now we must make cook up a situation where it's not even isomorphic to a strict monoidal functor.

John Baez (Mar 07 2022 at 19:04):

Now I'm curious about strict symmetric monoidal categories where the monoidal product is strict monoidal. I know commutative monoidal categories are of this sort - those are symmetric monoidal categories where the symmetry, or braiding, is the identity. But are there others?

John Baez (Mar 07 2022 at 19:05):

I ask because I know plenty of strict symmetric monoidal categories that are not equivalent commutative monoidal categories, like strictified versions of the category of finite sets, or finite-dimensional vector spaces.

Chad Nester (Mar 07 2022 at 19:10):

@Martti Karvonen Excellent, thanks!

Mike Shulman (Mar 07 2022 at 19:18):

John Baez said:

Now I'm curious about monoidal categories where the monoidal product is strict monoidal. I know commutative monoidal categories are of this sort - those are symmetric monoidal categories where the symmetry, or braiding, is the identity. But are there others?

Are you asking about symmetric monoidal categories or non-symmetric ones?

John Baez (Mar 07 2022 at 19:27):

Sorry, I actually meant strict symmetric ones - that's what's relevant here. I'll correct my comment.

Martti Karvonen (Mar 07 2022 at 19:33):

Shouldn't Eckman-Hilton imply that these are exactly the commutative ones?

Martti Karvonen (Mar 07 2022 at 19:38):

unless you meant "is monoidally isomorphic to a strict monoidal functor"

John Baez (Mar 07 2022 at 19:40):

I guess you're right, they're just the commutative ones.

John Baez (Mar 07 2022 at 19:42):

I was momentarily confused by the existence of medial semigroups that aren't commutative, where a semigroup is medial if it obeys (cd)(c'd') = (cc')(dd'). But a medial semigroup with an identity element is commutative (obviously), so this is irrelevant.

John Baez (Mar 07 2022 at 19:42):

I did not mean "monoidally isomorphic", I meant "equals".

John Baez (Mar 07 2022 at 19:44):

But I was leading up to a conjecture like this: if a symmetric monoidal category is not symmetric monoidal equivalent to a commutative one, it's not symmetric monoidal equivalent to one for which the tensor product is a strict monoidal functor.

Chad Nester (Mar 09 2022 at 13:32):

It's interesting that for props $\mathbb{X},\mathbb{Y}$ the category $\mathsf{StrSMC}(\mathbb{X},\mathbb{Y})$ doesn't seem to be symmetric monoidal in this way. In particular the object monoid of a prop is commutative for silly reasons and so we have that $F \otimes G$ as defined above is strict monoidal at the level of objects. However, the arrow monoid of a prop need not be commutative and so $F \otimes G$ fails to be a strict monoidal functor...

Chad Nester (Mar 09 2022 at 13:34):

I'm confused about this today: In Hackney and Robertson's paper they claim that the category of props is self-enriched, and we seem to be talking about the same things.

Chad Nester (Mar 09 2022 at 13:34):

Am I missing something, or is that not true?

Chad Nester (Mar 23 2022 at 12:36):

I'm convinced that there is indeed an error in "On the Category of PROPs" by Hackney and Robertson.

In the single-coloured case they state that their notion of PROP is equivalently a symmetric strict monoidal category generated by a single object (top of page 3), and their notion of PROP homomorphism (Definition 7) is pretty clearly a symmetric strict monoidal functor. Their category of PROP transformations $\mathsf{Hom}(\mathcal{R},\mathcal{T})$ defines the tensor product on objects pointwise in the manner discussed above.

It follows that their Proposition 24 -- that $\mathsf{Hom}(\mathcal{R},\mathcal{T})$ is a PROP, is in fact false, as discussed above. It also follows that the category of PROPs is not in fact self-enriched. Similarly, Proposition 38 and Theorem 39 do not in fact hold. Theorem 39 -- that the category of PROPs is monoidal closed -- is a major contribution of the paper, which has been cited 20 times. This is somewhat concerning.

Chad Nester (Mar 23 2022 at 12:40):

It doesn't seem like we can expect the situation with PROPs to be significantly simpler than the general one described by Schmitt.

Chad Nester (Mar 23 2022 at 12:44):

That is, the 2-category of PROPs, strong monoidal functors, and monoidal transformations ought to admit a "lax symmetric monoidal 2-categorical structure", and to be "lax closed" with respect to it.

Joe Moeller (Mar 23 2022 at 13:02):

You should email the authors. I know for a fact that Philip would like to chat about this. And he’s nice!

Chad Nester (Mar 23 2022 at 13:46):

That's a good idea :)

philip hackney (Apr 18 2022 at 14:54):

I chatted with @Chad Nester and I think we worked out what's going on here. The set of colors of the internal hom between two props is the set of prop maps. But if you regard this internal hom as a symmetric strict monoidal category, then its set of objects consists of finite ordered lists of prop maps. i.e. apply the free monoid monad to the set of colors to get the set of objects.

philip hackney (Apr 18 2022 at 14:54):

So one doesn't need to worry about evaluating the tensor product of two prop maps at an object or a morphism, since the tensor product of two prop maps isn't itself a prop map.

philip hackney (Apr 18 2022 at 14:58):

We didn't emphasize this point about the internal hom in the paper since we were usually thinking about props as props and not as symmetric strict monoidal categories.