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: terminology for halfway rigid monoidal categories

Chad Nester (Nov 10 2021 at 09:32):

A monoidal category is rigid in case every object therein has a left dual and a right dual.

I'm wondering if there's a name for monoidal categories in which every object has either a left dual or a right dual.

I'm tempted to go with semi-rigid, but if something already exists for this I'd like to use that. Also, has anyone else encountered categories like this? (more examples would be good!)

The "or" can be inclusive or exclusive here, although it's worth noting that in the exclusive case the monoidal unit is an exception.

Morgan Rogers (he/him) (Nov 10 2021 at 10:48):

You got an example?

Morgan Rogers (he/him) (Nov 10 2021 at 10:49):

(I think) duals on the same side have the nice property that $(A \otimes B)^* \cong B^* \otimes A^*$, so I'm curious what happens when you tensor objects which have duals on opposite sides.

Chad Nester (Nov 10 2021 at 10:55):

In this paper the "free cornering" of a symmetric (strict) monoidal category is a single object double category. If I consider the horizontal cells of that double category (those with trivial top and bottom boundary), I obtain a monoidal category. This category has the "semi-rigid" property. In the notation of the paper, $A^\circ$ is the left dual of $A^\bullet$ for all $A$ (and so $A^\bullet$ is the right dual of $A^\circ$), but $A^\circ$ need not have a left dual, and $A^\bullet$ need not have a right dual.

Chad Nester (Nov 10 2021 at 10:57):

This category of horizontal cells is odd, but might end up being important. Currently working on a journal-special-issue version of that paper which will include a discussion of its structure.

Nick Hu (Nov 10 2021 at 11:01):

'autonomous' is used as a synonym for 'rigid', and I've seen 'left autonomous' and 'right autonomous' to mean only left/right duals respectively (although never 'left rigid' etc.)

Maybe consult remark 4.2 of https://arxiv.org/abs/0908.3347

Nick Hu (Nov 10 2021 at 11:03):

2021-11-10-110249_667x150_scrot.png

Jules Hedges (Nov 10 2021 at 11:04):

I wonder if you can steal existing terminology from monoids? This thing is like a noncommutative monoid where everything has a left-inverse

Jules Hedges (Nov 10 2021 at 11:06):

(I like to think of autonomous categories as a categorification of groups - being a dual object is a better categorification of inverse elements than the obvious thing)

Morgan Rogers (he/him) (Nov 10 2021 at 11:31):

Jules Hedges said:

I wonder if you can steal existing terminology from monoids? This thing is like a noncommutative monoid where everything has a left-inverse

If everything has a left inverse, the monoid is a group!

Chad Nester (Nov 10 2021 at 12:28):

Is it? Every object has a left or right dual, but some objects don't have left duals, and some objects don't have right duals.

Chad Nester (Nov 10 2021 at 12:30):

The autonomous vs rigid thing isn't really the point. It's that instead of having both properties (left and right dual) for each object of the category, I have one of them.

Reid Barton (Nov 10 2021 at 12:31):

What happens when you tensor a left dual with a right dual, like Morgan mentioned before?

Chad Nester (Nov 10 2021 at 12:35):

Maybe nothing? All that comes to mind is that there's an interesting morphism $A^\circ \otimes B^\bullet \to B^\bullet \otimes A^\circ$ ($A^\circ$ is a left dual and $B^\bullet$ is a right dual). The morphism is interesting because there need not be a morphism in the other direction, which is what stops this from being a symmetric monoidal category.

Nathanael Arkor (Nov 10 2021 at 13:28):

I would have imagined that the fact that every object either has a left-dual or right-dual is less important than the fact that the category is equipped with a subcategory, whose objects all have right duals. That is, I would be surprised if there were theorems about "monoidal categories in which every object has a left or right dual" (which seems to me like a freeness property), but there are certainly theorems about monoidal categories that embed into monoidal categories in which every object has a right dual (these are one-object proarrow equipments), and so it may be that the latter is worth naming, but not the former.

Reid Barton (Nov 10 2021 at 13:33):

Does $A^\circ \otimes B^\bullet$ actually have either a left or right dual?

Chad Nester (Nov 10 2021 at 13:48):

@Nathanael Arkor I think you're right!

Chad Nester (Nov 10 2021 at 13:51):

Reid Barton said:

Does $A^\circ \otimes B^\bullet$ actually have either a left or right dual?

Embarassingly, no. My question isn't the right question to be asking about my example -- which isn't an example at all!

Morgan Rogers (he/him) (Nov 10 2021 at 15:59):

My curiosity is satisfied.

Jules Hedges (Nov 10 2021 at 16:02):

Morgan Rogers (he/him) said:

Jules Hedges said:

I wonder if you can steal existing terminology from monoids? This thing is like a noncommutative monoid where everything has a left-inverse

If everything has a left inverse, the monoid is a group!

I'm not convinced, can you write a proof? If this was true then every pregroup would collapse to just be a group, I guess...

Morgan Rogers (he/him) (Nov 10 2021 at 16:06):

Consider an element $x$. It has a left inverse $y$. Also, $y$ has a left inverse $z$. That is, $yx = 1 = zy$. But then $x = zyx = z$ is a two-sided inverse to $y$.

Mike Shulman (Nov 10 2021 at 17:45):

Doesn't "$x$ has a left inverse $y$" mean $y x = 1$, not $x y = 1$?

Morgan Rogers (he/him) (Nov 10 2021 at 17:53):

That seems reasonable, I have flipped everything accordingly.

John Baez (Nov 10 2021 at 17:57):

Morgan Rogers (he/him) said:

Jules Hedges said:

I wonder if you can steal existing terminology from monoids? This thing is like a noncommutative monoid where everything has a left-inverse

If everything has a left inverse, the monoid is a group!

Chad Nester said:

Is it?

Yeah, I just checked. Suppose you have a monoid where every element has a left inverse. Thus every $x$ has a left inverse, say $a$:

$a x = 1$

I claim $a$ is also the right inverse of $x$. This will prove Morgan's statement.

First, note that

$axa = a$

Now multiply on the left by the left inverse of $a$, which I'll call $b$:

$baxa = ba$

Since $ba = 1$ we get

$x a = 1$

so $a$ is the right inverse of $x$.

John Baez (Nov 10 2021 at 17:59):

(This is basically Morgan's argument above, but I wrote it before reading his latest comments. It was a fun puzzle.)

Nick Hu (Nov 10 2021 at 18:02):

Morgan Rogers (he/him) said:

Jules Hedges said:

I wonder if you can steal existing terminology from monoids? This thing is like a noncommutative monoid where everything has a left-inverse

If everything has a left inverse, the monoid is a group!

Does this cute result have a name by the way? I've seen it a couple times now

Mike Shulman (Nov 10 2021 at 18:10):

Another way to phrase the proof is: if $x$ has a left inverse $y$ and $y$ has a left inverse $z$, then $y$ has both a left and a right inverse. Therefore, those two coincide and are a two-sided inverse. This is no different from Morgan's argument, but it might make it sound more familiar.

Mike Shulman (Nov 10 2021 at 18:10):

It's also an instance of the [[2-out-of-6 property]]...

John Baez (Nov 10 2021 at 18:15):

Two out of six - cute!