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Stream: theory: category theory

Topic: commutative monoids in a braided monoidal cat

David Michael Roberts (May 28 2021 at 08:51):

(Ok, technically braided commutative monoids or something..)

So for a symmetric monoidal category $C$ the category of monoids in it is monoidal, and so is the category of commutative mononoids. I presume the same is true for the category of (commutative) monoids in a merely braided monoidal category? The only thing that I think is a little fiddly is making sure the product on the tensor product of (commutative) monoids uses the correct direction of the braiding, relative to the definition of "commutative".

What's a reference for this?

David Michael Roberts (May 28 2021 at 08:59):

Heh, Maxime Ramzi just answered this on Twitter, where I first asked it. The answer is no, for reasons of E_n algebra machinery.

Amar Hadzihasanovic (May 28 2021 at 09:04):

That's right. Btw the direction of the braiding does not matter since the (braided) theory of commutative monoids is self-dual with respect to inverting the direction of the braiding...

David Michael Roberts (May 28 2021 at 09:34):

I do wonder what exactly breaks: there's no product bifunctor on CommMon(C)? Or the coherence doesn't work? I'm happy to lose various bells and whistles. I'll have a bit more of a think....

Amar Hadzihasanovic (May 28 2021 at 11:12):

I wouldn't really know where to start for a “generic” no-go theorem (as in 'there is no braided monoidal structure at all', which may even not be true), but there are some “relative” no-gos for the kinds of structures that one would want.

To start, the category $\mathrm{Mon}(C)$ of monoids in a braided monoidal category $C$ does admit a monoidal structure -- in fact there are two, related by the “mirror symmetry” that inverts the direction of all braidings -- formed in the expected way: given monoids $(\mu_A, \eta_A)$ and $(\mu_B, \eta_B)$ their tensor product is given by $((\mu_A \otimes \mu_B) \circ (\mathrm{id}_A \otimes \tau_{B,A} \otimes \mathrm{id}_B), \eta_A \otimes \eta_B)$ or the same with $\tau_{A,B}^{-1}$ instead of $\tau_{B,A}$.

Amar Hadzihasanovic (May 28 2021 at 11:17):

However the “naive” attempt to give either of these a braided monoidal structure fails because the braidings $\tau_{A,B}$ are not going to be monoid homomorphisms from the tensor of two monoids to their tensor in the other order.

Amar Hadzihasanovic (May 28 2021 at 11:24):

These would be the only possibilities for a braided structure which is compatible with the “symmetrisation” $SC$ of $C$...

That is, more precisely, the inclusion of SMCs into BMCs has a left adjoint $S$; applying $\mathrm{Mon}(-)$ to the unit of the adjunction gives you a functor $\mathrm{Mon}(C) \to \mathrm{Mon}(SC)$ which is, in fact, a strong monoidal functor (with respect to either choice of monoidal structure on $\mathrm{Mon}(C)$, and the “standard” monoidal structure on $\mathrm{Mon}(SC)$).

Amar Hadzihasanovic (May 28 2021 at 11:27):

We know that $\mathrm{Mon}(SC)$ admits in fact a symmetric monoidal structure, so if $\mathrm{Mon}(C)$ had a braided monoidal structure, we would expect this to be also a braided monoidal functor; but the only possibility seems to be the “naive” one that I mentioned above and doesn't work. (This is a bit informal but I'm sure it can be made precise).

Amar Hadzihasanovic (May 28 2021 at 11:31):

So already $\mathrm{Mon}(C)$ does not have a braided structure with the properties that we would expect; restricting to commutative monoids does not change the situation...

David Michael Roberts (May 28 2021 at 11:59):

Cool, thanks for the analysis!

Amar Hadzihasanovic (May 28 2021 at 13:46):

Ah, I realised you didn't actually ask whether $\mathrm{Mon}(C)$ is braided, just whether $\mathrm{CommMon}(C)$ is monoidal! I don't know how I misunderstood the question.

Anyway, the answer then is that $\mathrm{CommMon}(C)$ is not closed under the tensor product of $\mathrm{Mon}(C)$ (as a full subcategory of it), with either choice of monoidal structure.