Timeline for Monoidal categories whose tensor has a left adjoint
Current License: CC BY-SA 4.0
13 events
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| Feb 3, 2021 at 16:17 | comment | added | Tim Campion | Well, I think Qiaochu did most of the work :) I also agree that even though the Day convolution argument appears not to work out, it is a great idea to try, and for those not familiar with such a technique, it's very instructive just to see an experienced category theorist reach for it. It illustrates the general method of "embed in a larger category with more structure" to solve one's problems. And the fact that such techniques exist helps clarify why it makes so much sense to do as Qiaochu did and simplify things by assuming the existence of extra structure at least to start. | |
| Jan 5, 2021 at 0:00 | comment | added | Qiaochu Yuan | @varkor: no worries, Tim should get much of the credit anyway! I always get worried when nobody bothers to actually check my arguments; I've had to fix gaps in my MO and math.SE answers that nobody pointed out for years... so I'm grateful to Tim for his corrections and elaborations! | |
| Jan 4, 2021 at 23:59 | comment | added | varkor | @QiaochuYuan: thank you! This is a very good answer (along with the argument via Day convolution that was here previously). It's tricky to know which answer to accept, because the process of developing the proof was very much a collaboration: I wish I could accept both answers. I've decided to accept Tim's answer, simply because it contains the full proof that $\mathscr V$ has finite products, but I appreciate that this answer already contains many of the key ideas. Sorry, and thank you again! | |
| Jan 4, 2021 at 17:19 | comment | added | Oscar Cunningham | I think this could be used to prove a version of the no-cloning theorem. If there was a way to clone quantum information then you could use it to define such an adjoint on $(\mathrm{Vect}_\mathbb{C},\otimes)$, with the $l$ and $r$ of the question being identities. Then $\otimes$ is a product, contradiction. | |
| Jan 3, 2021 at 22:51 | comment | added | varkor | Thanks, this is a very elegant argument. To check my understanding: if we require that the tensor and unit have left adjoints, then the unit will be terminal, and the argument demonstrating that $X \otimes Y \cong X \times Y$ carries through as written, establishing that finite products do exist in $\mathscr V$, without having to take it as an assumption, right? In this case, cartesian categories may be characterised as monoidal categories such that the tensor and unit are right adjoint. That the left adjoints are the diagonals follows automatically. | |
| Jan 3, 2021 at 22:14 | comment | added | Qiaochu Yuan | @Tim: nuts, you're right again. Thanks for checking the argument, I didn't feel 100% solid about it either. That's a nice observation about the unit also. | |
| Jan 3, 2021 at 22:12 | history | edited | Qiaochu Yuan | CC BY-SA 4.0 |
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| Jan 3, 2021 at 21:56 | comment | added | Tim Campion | @QiaochuYuan I'm not sure I follow the argument that the map $(X \times Y) \hat \otimes (Z \times W) \to ( X \hat \otimes Z) \times (Y \hat \otimes Z)$ is an isomorphism on $V$ -- the domain doesn't really simplify if finite products aren't known to exist in $V$. However, assuming that $1_\times$ exists, the isomorphism $X \otimes Y \cong X \times Y$ that you construct above doesn't require us to know that $X \times Y$ exists a priori -- only that $X \times 1$ and $1 \times Y$ do (which they do!). So maybe we don't need to think about Day convolution at all! | |
| Jan 3, 2021 at 21:45 | history | edited | Qiaochu Yuan | CC BY-SA 4.0 |
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| Jan 3, 2021 at 21:41 | comment | added | მამუკა ჯიბლაძე | Cannot really catch it but some kind of category of binary trees could provide a counterexample. I mean, $\ell$ and $r$ being left and right branch and $\otimes$ pruning (or how is it called) | |
| Jan 3, 2021 at 21:37 | comment | added | Tim Campion | Nice! In the last paragraph, what's not clear to me is whether $\hat \otimes$ inherits a left adjoint from $\otimes$ (so that the same argument can really be run with $\hat \otimes$). One tweak which might work better would be to work with the functor $Fam$ which freely completes under finite products. I believe that $Fam$ is 2-functorial (so that it preserves the adjunction) and preserves finite products (so that $Fam(\otimes)$ is a monoidal structure on $Fam(V)$ analogous to Day convolution). But the embedding $V \to Fam(V)$ doesn't preserve finite products, so perhaps I'm spouting nonsense. | |
| Jan 3, 2021 at 21:28 | history | edited | Qiaochu Yuan | CC BY-SA 4.0 |
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| Jan 3, 2021 at 21:01 | history | answered | Qiaochu Yuan | CC BY-SA 4.0 |