Category Theory
Zulip Server
Archive

You're reading the public-facing archive of the Category Theory Zulip server.
To join the server you need an invite. Anybody can get an invite by contacting Matteo Capucci at name dot surname at gmail dot com.
For all things related to this archive refer to the same person.

Stream: theory: category theory

Topic: monoidal closed but not compact closed

JS PL (he/him) (Dec 01 2021 at 16:51):

I was wondering if anyone had an example of a symmetric monoidal closed category where $A \multimap B = A \otimes B$ but which is not compact closed.

JS PL (he/him) (Dec 01 2021 at 16:53):

It's definitely true that a compact closed category where $A^\ast =A$ (like a hypergraph category, https://ncatlab.org/nlab/show/hypergraph+category#Carboni) is closed with $A \multimap B = A \otimes B$ but the converse is not true.

JS PL (he/him) (Dec 01 2021 at 16:54):

What's missing is that a certain canonical map is not necessarily an isomorphism (see nuclear objects https://ncatlab.org/nlab/show/nuclear+object)

JS PL (he/him) (Dec 01 2021 at 16:55):

I can always synthetically build the free one via some prop, but I was wondering if someone had another example in mind.

Mike Shulman (Dec 01 2021 at 17:17):

The converse is not obviously true, but without a counterexample how do you know that it doesn't happen to be true accidentally?

JS PL (he/him) (Dec 01 2021 at 17:24):

Fair point!

JS PL (he/him) (Dec 01 2021 at 17:25):

If it is accidentally true, I would love to see a proof! I've been playing around with this for while and can't seem to prove the snake equations.

Mike Shulman (Dec 01 2021 at 17:25):

I would be surprised if it's accidentally true, but I've been surprised before.

JS PL (he/him) (Dec 01 2021 at 17:26):

My other intiuition is that the free such monoidal closed category will not be compact closed.

JS PL (he/him) (Dec 01 2021 at 17:29):

Mike Shulman said:

I would be surprised if it's accidentally true, but I've been surprised before.

Essentially my understanding is this:
In such a monoidal closed category you would have the evaluation map $A \otimes A \otimes B \xrightarrow{ev} B$

Mike Shulman (Dec 01 2021 at 17:30):

Yes, it feels for instance like there should be $\ast$-automonous categories in which $\otimes = ⅋$ but $1\neq \bot$.

JS PL (he/him) (Dec 01 2021 at 17:32):

To get the snake equations you would need that this evaluation map factors through evaluation of $A$ and the monoidal unit $I$, that is,
$A \otimes A \otimes B \xrightarrow{\cong} A \otimes A \otimes I \otimes B \xrightarrow{ev \otimes B} I \otimes B \xrightarrow{\cong} B$.

JS PL (he/him) (Dec 01 2021 at 17:32):

But in general I don't see why such an equality would hold.

JS PL (he/him) (Dec 01 2021 at 17:34):

Mike Shulman said:

Yes, it feels for instance like there should be $\ast$-automonous categories in which $\otimes = ⅋$ but $1\neq \bot$.

Doesn't $\otimes = ⅋$ imply $1 = \bot$? Or at least $1 \cong \bot$?

JS PL (he/him) (Dec 01 2021 at 17:36):

$1 \cong \bot ⅋ 1 = \bot \otimes 1 \cong \bot$

Mike Shulman (Dec 01 2021 at 17:39):

Ah yes, that's right! (-:O

Mike Shulman (Dec 01 2021 at 17:43):

You know, actually I think this should be true. If $A\multimap B \cong A\otimes B$, then for any $X$ you have a natural isomorphism $\hom(A\otimes X,B) \cong \hom(X, A\otimes B)$, so we have an adjunction $(A\otimes -) \dashv (A\otimes -)$. I think that implies that $A$ is dual to itself.

Mike Shulman (Dec 01 2021 at 17:43):

Just take the unit and counit of that adjunction and evaluate them at the unit object.

Martti Karvonen (Dec 01 2021 at 17:50):

I remember seeing an old discussion on the cafe about whether an adjunction $(A\otimes -)\dashv (B\otimes -)$ is sufficient for $A$ being the dual of $B$, and I think the conclusion was no. I can try to find it if needed.

Martti Karvonen (Dec 01 2021 at 17:55):

remark 2.3 at [[dualizable object]] alludes to this but doesn't give an example

Mike Shulman (Dec 01 2021 at 17:56):

Oh, I think I remember that now. I think I make this mistake a lot. Thanks...

Mike Shulman (Dec 01 2021 at 17:56):

It would be nice to have a concrete counterexample though.

Matteo Capucci (he/him) (Dec 01 2021 at 18:19):

Martti Karvonen said:

I remember seeing an old discussion on the cafe about whether an adjunction $(A\otimes -)\dashv (B\otimes -)$ is sufficient for $A$ being the dual of $B$, and I think the conclusion was no. I can try to find it if needed.

That's the definition of 'dual' that Street gives, iirc

Mike Shulman (Dec 01 2021 at 18:21):

Where?

Matteo Capucci (he/him) (Dec 01 2021 at 18:22):

I don't remember the name of the paper, I only remember it's handwritten and about compact closed categories

Matteo Capucci (he/him) (Dec 01 2021 at 18:23):

Let me see if I find it

Matteo Capucci (he/him) (Dec 01 2021 at 18:24):

Here, p.38

JS PL (he/him) (Dec 01 2021 at 18:37):

Mike Shulman said:

It would be nice to have a concrete counterexample though.

Yes it would!

Mike Shulman (Dec 01 2021 at 18:37):

Thanks. It looks to me like at the bottom of p37 he wrote "The adjointness $A\dashv B$ between objects is in fact equivalent to an adjointness $-\otimes B \dashv -\otimes A$ between functors", but then crossed out "is in fact equivalent to" and replaced it with "implies".

John Baez (Dec 01 2021 at 20:29):

Ha!

John Baez (Dec 01 2021 at 20:30):

It's true!

John Baez (Dec 01 2021 at 20:32):

Maybe he pulled down those notes, made the correction and rescanned that page yesterday. :upside_down:

Graham Manuell (Dec 02 2021 at 16:12):

Martti Karvonen said:

I remember seeing an old discussion on the cafe about whether an adjunction $(A\otimes -)\dashv (B\otimes -)$ is sufficient for $A$ being the dual of $B$, and I think the conclusion was no. I can try to find it if needed.

Mike Shulman said:

It would be nice to have a concrete counterexample though.

I also remembered reading this. It is difficult to google, but luckily I seems that I had thought to download it and so I can give the link: https://golem.ph.utexas.edu/category/2008/02/logicians_needed_now.html#c018567.

Mike Shulman (Dec 02 2021 at 17:47):

Thanks for that link! Can it be made even more explicit for those of us who aren't familiar with fusion categories, and so for whom the conclusion $1 = \sum_k N_{ij}^k$ is not "clearly" wrong? Can you describe an explicit particular fusion category for which this equality fails? (Also it would be nice to see the "calculation best done in string diagrams" written out.)

Mike Shulman (Dec 02 2021 at 17:49):

Also, the original question on this thread was looking for a category that is not compact closed. If I understand correctly, a fusion category is compact closed (or at least left and right autonomous — maybe it's not symmetric?), and the counterexample is just a particular object that isn't a dual even though there is also an actual dual object. Can this be improved to an example of an object satisfying this weaker condition but that doesn't have any dual?

Graham Manuell (Dec 02 2021 at 19:51):

I can't help here unfortunately, because I don't really know about fusion categories either. I agree that an example where the object isn't dualisable at all would be nice. Maybe it's possible to pass to some subcategory of that one that doesn't contain the dual? Hopefully someone else will be able to help us understand the details.

Matteo Capucci (he/him) (Dec 03 2021 at 09:24):

Mike Shulman said:

Thanks. It looks to me like at the bottom of p37 he wrote "The adjointness $A\dashv B$ between objects is in fact equivalent to an adjointness $-\otimes B \dashv -\otimes A$ between functors", but then crossed out "is in fact equivalent to" and replaced it with "implies".

Uhm, right. Does it mean $A \otimes - \vdash B \otimes -$ does not imply $A = B^*$ in general?

JS PL (he/him) (Dec 03 2021 at 13:17):

Matteo Capucci (he/him) said:

Uhm, right. Does it mean $A \otimes - \vdash B \otimes -$ does not imply $A = B^*$ in general?

Yes that's the conclusion. The counter-example can be found in the link @Graham Manuell shared above.

John Baez (Dec 03 2021 at 19:43):

Right, we don't necessarily get $B \cong A^\ast$.

Matteo Capucci (he/him) (Dec 06 2021 at 21:23):

JS Pacaud Lemay (he/him) said:

Matteo Capucci (he/him) said:

Uhm, right. Does it mean $A \otimes - \vdash B \otimes -$ does not imply $A = B^*$ in general?

Yes that's the conclusion. The counter-example can be found in the link Graham Manuell shared above.

Oh, yeah. Lately I'm really prone to blindness.

JS PL (he/him) (Jan 18 2022 at 12:47):

So I still haven't found the counterexample I'm looking for... (the monoidal closed example where $A \multimap B = A \otimes B$ but which is not compact closed)

JS PL (he/him) (Jan 18 2022 at 12:54):

But I was wondering if anyone had a counterexample for the following:
In a symmetric monoidal closed category, a nuclear object (https://nlab-pages.s3.us-east-2.amazonaws.com/nlab/show/nuclear+object) is an object $A$ such that the canonical map $A \otimes (A \multimap I) \to A \multimap A$ is an isomorphism (where $I$ is the unit)

JS PL (he/him) (Jan 18 2022 at 12:55):

Every nuclear object $A$ has a dual $A^\ast = A \multimap I$

JS PL (he/him) (Jan 18 2022 at 12:56):

I'm curious about the converse: if $A$ has a dual $A^\ast$, is it nuclear?

JS PL (he/him) (Jan 18 2022 at 12:57):

Here it says that this might be the case: https://nlab-pages.s3.us-east-2.amazonaws.com/nlab/show/dualizable+object (see section 4) -- but I get stuck on the proof... so I went looking for a reference that nuclear = dualizable, but this does not seem to be the case? So then I was looking for a counter-example and couldn't find one!

JS PL (he/him) (Jan 18 2022 at 12:58):

Just wondering if by chance someone had an answer. Though nuclear objects don't seem to have been studied much in recent years...

JS PL (he/him) (Apr 04 2022 at 06:35):

Mike Shulman said:

Yes, it feels for instance like there should be $\ast$-automonous categories in which $\otimes = ⅋$ but $1\neq \bot$.

@Mike Shulman out of curiosity, do you have a simple example of a $\ast$-autonomous category where $\otimes = ⅋$? (or even just a linear distributive category...)

Mike Shulman (Apr 04 2022 at 06:37):

Any compact closed category is $\ast$-autonomous with $\otimes = ⅋$ and $1=\bot$. Any symmetric monoidal category is likewise degenerately linearly distributive.

JS PL (he/him) (Apr 04 2022 at 06:39):

Ah yes sorry I should have been more precise: I was hoping for one that wasn't compact closed :upside_down:

JS PL (he/him) (Apr 04 2022 at 06:39):

but that might be asking for too much because that would solve my problem I guess...

Mike Shulman (Apr 04 2022 at 06:51):

I don't think I know any examples of that.