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Stream: learning: questions

Topic: Equivalent characterizations of compact closed category

Samuel Steakley (Aug 13 2024 at 03:43):

In "Coherence for Compact Closed Categories," Kelly and Laplaza introduce compact closed categories by the following definition:

a symmetric monoidal closed category whose internal hom $[A,C]$ has the form $C\otimes A^*$.

They explain how this definition implies the existence of duals: for $A$ to be dualizable, it suffices that the "canonical map" $A\otimes [A,I] \to [A,A\otimes I]$ be an isomorphism (I understand this canonical map to be the adjunct of $1_A\otimes \textup{ev}_I$). They seem to suggest that this map being an isomorphism should follow from their initial definition of compact closed category, but I haven't figured out how to prove it yet. Any tips or references that might help to prove this fact?

John Baez (Aug 13 2024 at 08:11):

I don't think one can do much simply knowing that $[A,C]$ is "of the form" $C \otimes A^\ast$ - if by "of the form" one merely means "isomorphic" or even "naturally isomorphic". One needs a natural isomorphism that obeys some properties.

Kelly is a very careful guy. So I don't think

a symmetric monoidal closed category whose internal hom $[A,C]$ has the form $C\otimes A^*$.

is his official definition of compact closed category.

I bet in this passage he's either just giving you a vague reminder of the concept of compact closed category (which you are supposed to know already), or making a vague preliminary remark to help you understand the basic idea (which he will make precise later).

Samuel Steakley (Aug 13 2024 at 08:20):

Thank you, that's very helpful.

In this paper, they go on to adopt the definition in terms of dual objects. Any idea where I might find discussion of an equivalent definition based on the closed structure?

John Baez (Aug 13 2024 at 08:21):

Okay - I looked at the paper, an it's not quite as clear as I'd like: they don't really give a definition of compact closed category before they "analyze" the definition. Then later they give an actual definition of compact closed category on the second page, where they say "We can therefore adopt the following simple definition of compact closed category..."

John Baez (Aug 13 2024 at 08:32):

You can also define compact closed categories this way.

A symmetric monoidal category $\mathsf{C}$ is closed if for each object $A \in \mathsf{C}$ the functor $- \otimes A : \mathsf{C} \to \mathsf{C}$ has a right adjoint. That means there's an isomorphism, natural in $B \in \mathsf{C}^{\rm op}, C \in \mathsf{C}$, like this:

$\mathsf{C}(B \otimes A, C) \cong \mathsf{C}(B, [A,C])$

You can say a symmetric monoidal category $\mathsf{C}$ is compact closed if for every object $A$ there is an object $A^\ast$ such that there's an isomorphism, natural in $B \in \mathsf{C}^{\rm op}, C \in \mathsf{C}$, like this:

$\mathsf{C}(B \otimes A, C) \cong \mathsf{C}(B, C \otimes A^\ast)$

John Baez (Aug 13 2024 at 08:37):

Note with this definition any compact closed category is automatically closed, with $[A,C] \cong C \otimes A^\ast$. But the definition says more than just that.

John Baez (Aug 13 2024 at 08:38):

I hope I got everything right here. Stripping down these definitions to their bare minimum, without going too far and leaving out something essential, can be a bit tricky.

John Baez (Aug 13 2024 at 08:40):

I don't know who first introduced the definition of compact closed category! The most common definition nowadays is the one that Kelly and Laplaza call "the following simple definition..."

Samuel Steakley (Aug 13 2024 at 08:44):

John Baez said:

You can say a symmetric monoidal category $\mathsf{C}$ is compact closed if for every object $A$ there is an object $A^\ast$ such that there's an isomorphism, natural in $B \in \mathsf{C}^{\rm op}, C \in \mathsf{C}$, like this:

$\mathsf{C}(B \otimes A, C) \cong \mathsf{C}(B, C \otimes A^\ast)$

Ah, I do believe your definition is essentially the one I was using to try and attack my proposition. But! Re-reading it here makes me realize I may have been overlooking a certain fact about it. I will give it another try....

Samuel Steakley (Aug 13 2024 at 08:46):

John Baez said:

I don't know who first introduced the definition of compact closed category!"

Me neither, and I wish I did! Before now, I have only seen the definition in terms of duals.

John Baez (Aug 13 2024 at 08:49):

You'll notice that both in my definition of 'closed' and 'compact closed' I didn't say anything about the isomorphism being natural in $A \in \mathsf{C}^{\rm{op}}$. But this should follow.

More precisely, it should follow from the definitions I gave that $[A,C]$ and $A^\ast$ are contravariantly functorial in $A$, and that the isomorphisms

$\mathsf{C}(B \otimes A, C) \cong \mathsf{C}(B, [A,C])$

and

$\mathsf{C}(B \otimes A, C) \cong \mathsf{C}(B, C \otimes A^\ast)$

are natural in $A \in \mathsf{C}^{\rm op}$.

John Baez (Aug 13 2024 at 08:50):

This takes some work to show!

Samuel Steakley (Aug 13 2024 at 08:51):

Indeed, and I am grateful to have Mac Lane to cite for this instead of proving it myself :)

Todd Trimble (Aug 15 2024 at 01:01):

We once had a long conversation (including me and Robin Houston and Bruce Bartlett and Jamie Vicary and John Baez) about related matters at the n-Category Cafe. One may define a compact closed category as a symmetric monoidal category in which every object $C$ has a monoidal dual $C^\ast$ (left or right dual, take your pick; since the monoidal product is symmetric monoidal, it doesn't matter which). Now, speaking from personal experience, it can be tempting to say that $C^\ast$ is a monoidal dual of $C$ if $C^\ast \otimes -$ is right adjoint to $C \otimes -$, i.e., if there is a natural isomorphism

$\mathcal{C}(A, C^\ast \otimes B) \cong \mathcal{C}(C \otimes A, B).$

But this is not enough! You need an extra condition to get the usual notion of monoidal dual. If $c_X: C \otimes C^\ast \otimes X \to X$ denotes a typical component of the counit of $C \otimes - \dashv C^\ast \otimes -$, then you need also

$c_{I \otimes X} = c_I \otimes 1_X$

(see the comment here).

Eventually it was Bruce Bartlett who explained that this condition really doesn't come for free, in a comment here.

It's a fairly subtle issue. See also Remark 2.16 here.

Samuel Steakley (Aug 15 2024 at 01:16):

Ah, wonderful. This extra condition you mention happens to be exactly what Robin Cockett and I hit upon while discussing the question yesterday, but we weren't able to resolve it on the spot. Thanks very much Todd!

Amar Hadzihasanovic (Aug 15 2024 at 07:22):

One conceptual way of understanding the characterisation at the beginning is to consider it a special case of a result in formal 2-category theory, by translating facts about a symmetric monoidal category $\mathcal{C}$ into facts about its delooping $\mathrm{B}\mathcal{C}$, a 2-category with a single object.
In the latter, the 1-morphisms - which are all endomorphisms of the unique object - are the objects of $\mathcal{C}$ and the 2-morphisms are the morphisms of $\mathcal{C}$.

Amar Hadzihasanovic (Aug 15 2024 at 07:27):

Then, an internal hom $[A, B]$ in $\mathcal{C}$ with its evaluation map $\mathrm{ev}_{A,B}: A \otimes [A, B] \to B$ is the same as a right Kan extension of $B$ along $A$ in $\mathrm{B}\mathcal{C}$.

So the following are equivalent:

• $\mathcal{C}$ is closed,
• $\mathrm{B}\mathcal{C}$ has right Kan extensions of all morphisms along all morphisms.

Amar Hadzihasanovic (Aug 15 2024 at 07:32):

Now, in a 2-category, a right Kan extension $\rho: \mathrm{Ran}_p f \circ p \Rightarrow f$ of $f$ along $p$ is preserved by a morphism $g$ if $g \circ \rho$ is a right Kan extension of $g \circ f$ along $p$.

A right Kan extension is called absolute if it is preserved by every morphism.

Amar Hadzihasanovic (Aug 15 2024 at 07:36):

So in $\mathrm{B}\mathcal{C}$, an absolute right Kan extension is the same thing as an evaluation map $\mathrm{ev}_{A, B}$ with the property that, for all $C$, $\mathrm{ev}_{A, B} \otimes C$ is an evaluation map itself.

Which translates to the fact that the canonical morphism $[A, B] \otimes C \to [A, B \otimes C]$ is an isomorphism.

Amar Hadzihasanovic (Aug 15 2024 at 07:39):

Now, $A$ has a dual in $\mathcal{C}$ if and only if $A$ is (let's say right) adjoint as a 1-morphism in $\mathrm{B}\mathcal{C}$.

Amar Hadzihasanovic (Aug 15 2024 at 07:41):

There is a general result in formal 2-category theory which says that, for a 1-morphism $f: x \to y$, the following are all equivalent:

• $f$ is a right adjoint,
• a right Kan extension of $\mathrm{id}_x$ along $f$ exists and is absolute,
• a right Kan extension of $\mathrm{id}_x$ along $f$ exists and is preserved by $f$ itself.

Amar Hadzihasanovic (Aug 15 2024 at 07:45):

It is clear, moreover, that if an absolute right Kan extension of $\mathrm{id}_x$ along $f$ exists, then an absolute right Kan extension of any morphism along $f$ exists (just whisker the first one with that morphism!).

Amar Hadzihasanovic (Aug 15 2024 at 07:46):

So the following are equivalent:

• $\mathcal{C}$ is compact closed,
• $\mathrm{B}\mathcal{C}$ has absolute right Kan extensions of all morphisms along all morphisms.

Amar Hadzihasanovic (Aug 15 2024 at 07:48):

But by the result above, the latter is equivalent to

• for all objects $A$ of $\mathcal{C}$, $\mathrm{B}\mathcal{C}$ has a right Kan extension of $I$ along $A$, which is preserved by $A$.

Amar Hadzihasanovic (Aug 15 2024 at 07:50):

So if we assume that $\mathcal{C}$ is closed to begin with, then to prove that it is compact closed, it suffices to show that $\mathrm{ev}_{A, I} \otimes A$ is an evaluation map for all $A$, which, as we saw above, is equivalent to the canonical map $[A, I] \otimes A \to [A, A \otimes I]$ being an isomorphism.

Amar Hadzihasanovic (Aug 15 2024 at 07:50):

Which is the Kelly-Laplaza characterisation mentioned at the beginning!

Amar Hadzihasanovic (Aug 15 2024 at 07:56):

(By the way, all of this works in the non-symmetric case as well, giving a characterisation of the existence of one-sided duals; one has to dualise from right Kan extensions to right Kan lifts to get the other dual.)

John Baez (Aug 15 2024 at 09:10):

Very interesting stuff, @Amar Hadzihasanovic! Are there good references for these characterizations of closure or compact closure in terms of Kan extensions in the delooping of $\mathcal{C}$? I'd like to add this material to the nLab (though I'd be even happier if you did it).

John Baez (Aug 15 2024 at 09:12):

While working on the nLab page [[compact closed category]], I met this claim:

The [[delooping]] $\mathbf{B}M$ of a [[commutative monoid]] $M$ is a compact closed category...

This seems wrong to me. I thought $\mathbf{B}M$ was compact closed only when $M$ is a group. Can someone comment on this?

Amar Hadzihasanovic (Aug 15 2024 at 09:25):

I think whoever wrote that had in mind what I'd rather call the double delooping, so the monoidal category with only the monoidal unit as object, and $M$ as the commutative monoid of endomorphisms of the unit.

John Baez (Aug 15 2024 at 09:31):

Oh! That's better.

John Baez (Aug 15 2024 at 10:39):

How about my first question?

John Baez said:

Very interesting stuff, Amar Hadzihasanovic! Are there good references for these characterizations of closure or compact closure in terms of Kan extensions in the delooping of $\mathcal{C}$? I'd like to add this material to the nLab (though I'd be even happier if you did it).

Amar Hadzihasanovic (Aug 15 2024 at 10:59):

Yes, sorry, I've been trying to think of a reference but I don't think I know one. The 2-categorical stuff is classical and covered in textbooks, e.g. in the "All concepts are Kan extensions" chapters either in Mac Lane or Riehl, and the fact that "internal homs = right Kan extensions/lifts in the delooping" is one of those things that are treated as either obvious or "abstract nonsense for its own sake" so rarely put in print, if ever.

John Baez (Aug 15 2024 at 10:59):

Thanks.