The n-Category Café

Todd wrote:

Sorry – you already did mention in TWF 268 that last point about *-autonomous categories being Frobenius pseudomonoids in the bicategory of categories and profunctors.

Yes, this is the result I was attempting to give people a feeling for. But since normal folks run away screaming when someone says “a *-autonomous category is just a Frobenius pseudomonoid in the bicategory of categories and pseudofunctors”, I wanted to sneak up on this idea very, very cautiously.

So, I wanted to first sketch how *-autonomous categories naturally show up in logic, and then segue over to Street’s description of *-autonomous categories as Frobenius pseudomonoids.

But, I’m also trying to understand Melliès’ work on ‘game semantics’, so I wanted to work that in too…

I know that Melliès was very much influenced by your paper with Gerry Brady. So, he must know this result of yours, and he must have been trying to explain it to me:

… the notion of *-autonomous category could be expressed in exactly these terms: a symmetric monoidal $C$ equipped with a strong contravariant adjoint equivalence $\neg :C\to C$ [where the unit and counit of the adjunction respect the strengths].

But, the philosophy behind ‘game semantics’ seems to involve treating the two copies of $C$ as two ‘players’ — the guy who is fighting to prove some proposition, and the challenger who is trying to disprove them! The de Morgan duality built into logic is supposed to be nicely captured by this notion. To bring out this idea, Melliès prefers to talk about $C$ versus $C^{\mathrm{op}}$, instead of two copies of $C$.

This should be a minor esthetic decision, at least for defining $*$-autonomous categories… so I should be able to take your result and formulate it in terms of a covariant adjoint equivalence between $C$ and $C^{\mathrm{op}}$. Right?

It’s supposed to be like defining a Frobenius algebra by saying you have an algebra $A$, another algebra structure on $A^{*}$, and an isomorphism between them with some nice properties.

But this would be incredibly shocking

Sorry, I’m not sure whether you mean you believe me or don’t believe me! :-) Right now I’m leaning towards the former, but for what it’s worth, I overcame my earlier laziness and worked through the details, and I believe my earlier self! :-)

Sorry for the ambiguity. I was trying to say it would be incredibly shocking if we could not lift an adjoint equivalence to a monoidal adjoint equivalence in the situation you described.

I tried to explain my intuition: obviously we can lift the adjoint equivalence if it’s an isomorphism, and “I bet an adjoint equivalence is as close to an isomorphism of categories as we can get without going ‘evil’ and imposing equations between objects.”

In other words: I bet that given any “non-evil” structure on a category $C$, and an adjoint equivalence $f:C\to D$, we can use $f$ to transfer this structure to $D$ and then lift $f$ to an adjoint equivalence respecting this extra structure.

And then I mused that it would be interesting to try to prove a precise theorem along these lines.

Doing this would require more formally understanding structures on categories that you can define using commutative diagrams that only impose equations at the 2-morphism — natural transformation — level. These are the ‘non-evil’ structures.

A typical example of a non-evil structure is ‘weak monoidal category’. A typical example of an evil one is ‘strict monoidal category’.

For any non-evil structure $S$ there should be a 2-category ${\mathrm{Cat}}_S$ of categories equipped with this structure, functors weakly but coherently preserving this structure, and natural transformations compatible with this structure.

We know how ${\mathrm{Cat}}_S$ is defined when $S$ is the structure ‘monoidal category’. Is there an automatic way to define ${\mathrm{Cat}}_S$ starting from $S$? That’s the tricky part. But, I can imagine someone has already tackled it.

If this works, we get a forgetful 2-functor

$$F:{\mathrm{Cat}}_S\to \mathrm{Cat}$$

And then, we can try to prove that given a category $C\in \mathrm{Cat}$, and a way of equipping it with this extra structure, say $\tilde{C}\in {\mathrm{Cat}}_S$ with

$$F(\tilde{C})=C$$

and an adjoint equivalence between $C$ and $D$, we can lift this adjoint equivalence to one in ${\mathrm{Cat}}_S$.

Clearly this is not the quick way to prove the result we’re talking about.

James wrote:

Is there a lesson to be learned from the difference between adjoint equivalences and equivalences? Unless I’m mistaken, the definition of equivalence is officially non-evil, but it’s still not the best thing. The weird thing for me is that I’ve always liked the concept of adjoint equivalence better than that of equivalence […] but short of analyzing the homotopy type of their walking versions, I still can’t put my finger on why it’s better.

We don’t really need to take homotopy types of nerves to say what’s going on here. We can say it a different way — which may or may not make you feel more enlightened.

Namely: there’s a little category called ‘the walking isomorphism’: two objects and an isomorphism between them. This category is equivalent to ‘the walking object’: the category with just one object and its identity morphism.

This is a powerfully precise way of saying that “having two things with an isomorphism between them is just like having one thing”.

To see this, let $1$ be the walking object and let $\mathrm{Iso}$ be the walking isomorphism. Then our equivalence

$$1\simeq \mathrm{Iso}$$

automatically gives an equivalence

$$\mathrm{hom}(1,C)\simeq \mathrm{hom}(\mathrm{Iso},C)$$

for any category $C$. Of course

$$\mathrm{hom}(1,C)\simeq C$$

On the other hand, $\mathrm{hom}(\mathrm{Iso},C)$ is the category of ‘pairs of objects in $C$ equipped with an isomorphism between them’.

So, the category of ‘pairs of objects in $C$ equipped with an isomorphism between them’ is equivalent to $C$.

This nicely formalizes the idea that having two things with an isomorphism between them is just like having one thing!

On the other hand, look what happens when we go up to 2-categories.

Let $1$ be the 2-category with one object, its identity morphism, and its identity 2-morphism. And let $\mathrm{Equiv}$ be the walking equivalence: the 2-category consisting of two objects and an equivalence between them.

We might think that $1$ was equivalent to $\mathrm{Equiv}$, but it’s not true!

So, when we’re in a 2-category — e.g. Cat — having two things with an equivalence between them is not just like having one thing.

On the other hand, let $\mathrm{AdEquiv}$ be the ‘walking adjoint equivalence’. Then we have

$$1\simeq \mathrm{AdEquiv}$$

so for any 2-category

$$C\simeq \mathrm{hom}(1,C)\simeq \mathrm{hom}(\mathrm{AdEquiv},C)$$

So: having two things and an adjoint equivalence between them is just like having one thing.

Technical note: here I’m using the sensibly weakened concept of equivalence between 2-categories, which some people call ‘biequivalence’.

Puzzle: I said that the equivalence

$$C\simeq \mathrm{hom}(\mathrm{Iso},C)$$

meant having one object in $C$ was just like having two objects in $C$ and an isomorphism between them. But later I said the concept of equivalence was in some sense defective: having two things and an equivalence between them is not just like having one thing!

So: is having $C$ and $\mathrm{hom}(\mathrm{Iso},C)$ just like having $C$, or not?

Thanks. That’s interesting, but I guess I hoping for something closer to the syntactic level, something along the lines of “First make all existential quantifiers part of the structure, and then if you can state a relation in terms of your structure, you should”.

For example, what does it mean for $C$ to be equivalent to $D$? One definition is that we have two functors $f,g$ such that $\mathrm{fg}\cong 1_C$ and $\mathrm{gf}\cong 1_D$. But there are $\exists$s hiding in the $\cong$s, so we should really add structure. Then we get the data of a map $\alpha :\mathrm{fg}\to 1_C$, a map $\beta :\mathrm{gf}\to 1_D$, and their inverses ${\alpha }^{-1}$ and ${\beta }^{-1}$. So now we’ve converted what used to be properties on $f$ and $g$ into structure. But now that we have that structure, it’s possible to state the triangle axioms, so (by the principle above) we should! Then we have the concept of adjoint equivalence.

Is this principle a reasonable way of understanding why equivalence is not as good as adjoint equivalence? (It seems very close to the requirement that all homotopies be contractible.) If so, what is its most natural general formulation?

James wrote:

That’s interesting, but I guess I hoping for something closer to the syntactic level, something along the lines of “First make all existential quantifiers part of the structure, and then if you can state a relation in terms of your structure, you should”.

Okay. This motto seems to consist of two separate parts. I was focusing on this part:

“…then if you can state a relation in terms of your structure, you should.”

In topology this shows up in the recipe for ‘killing homotopy groups’, which says: “whenever you see a map from $S^j$ into your space $X$, use this to glue on an $(j+1)$-disk.” This makes the $j$th homotopy group of your space trivial. It creates an enormous $(j+1)$st homotopy group, which you can then go ahead and kill by the same method, if you want.

Translated into $n$-category language, this recipe says “whenever you see two $j$-morphisms $f,g:x\to y$, throw in an $(j+1)$-morphism $\alpha :f\to g$.” We can do this starting with whatever $j$ we want and working our way up. When we’re working with $n$-categories, and $j$ reaches $n$, a $(j+1)$-morphism is the same as an equation, and we can stop there.

(All equations between two fixed $n$-morphisms are automatically equal — so all the higher-dimensional holes are filled in by definition.)

All this sounds a bit complicated, but it’s actually a way of making an $n$-categorical structure as boring as possible. We do it when we don’t want complications: when we want a bunch of things to be equivalent whenever possible. This is the sense in which an adjoint equivalence is less complicated than an equivalence.

The other part of your motto:

“First make all existential quantifiers part of the structure…”

is another important aspect of the category-theorists’ worldview, but perhaps logically separate. Simply put: knowing that something exists is not nearly as useful as having it in your hands.

Somehow combining these mottos can lead us to the definition of adjoint equivalence.

By the way — someone should try my puzzle!

Good. Thanks. I learned about the $\exists$-removing principle a long time ago but didn’t know about the state-whatever-axioms-as-you-can principle. It’s tempting to call old-fashioned equivalence an under-achiever, what with it just sitting there leaving extra axioms unstated. I’m also tempted to say that “equivalence” should be redefined to be adjoint equivalence. This shouldn’t be a problem, because they were the same anyway in CAT. Maybe 2-category-theorists would object, but hey, they probably still call themselves bicategory-theorists.

By the way, do you know when this distinction between equivalence and adjoint equivalence was first noticed?

Regarding the puzzle, I’m not sure I understand the question. Are you asking whether $C$ is adjoint equivalent to $\mathrm{hom}(\mathrm{Iso},C)$?

You indeed want the comultiplication to be isometric i.e. δ^† o δ - or ‘special’ as it is sometimes called. I tend to forget to mention that since in QM terms this merely means physical realisability - if one believes ‘unitary = realisable’.

Bruce wrote:

Perhaps I am not on the same boat as far as what the morphisms are required to be: I am working in the paradigm where a morphism of commutative dagger Frobenius algebras is defined as a morphism of Frobenius algebras (i.e. not just a morphism of algebras).

Do you know the cute string diagram proof that a morphism of Frobenius algebras (i.e. not just their underlying algebras) is automatically an isomorphism?

I ask because I only learned it recently, and found it surprising at first. That’s why I mentioned it in week268.

So, your paradigm gives a groupoid of commutative dagger Frobenius algebras. To get the category of finite sets starting from Frobenius algebras, one needs another paradigm.

Frobenius algebras are defined by a PROP that happens to be compact (viewed as a symmetric monoidal category). I hear that for any sort of algebraic gadget defined by a PROP that is compact, the category of these gadgets is actually a groupoid. I haven’t checked this yet.

I’m not sure if this is relevant, but given a complete set of idempotents, i.e.

$$\sum _i{e}_i=1,$$

we can define a nice abstract graded differential algebra on them.

From there, we could construct a graph whose nodes are $e_i$ and whose edges are defined by

$$e_{\mathrm{ij}}=e_i{\mathrm{de}}_j{e}_j.$$

This gives a noncommutative discrete geometry on the algebra, which might tell you something.

Bruce said:

My understanding is that to specify a commutative daggger Frobenius algebra, one needs to specify the idempotents $e_i$ and the scale factors ${\lambda }_i=g(e_i,e_i)$ (if the algebra is separable, these factors are unity)

This isn’t quite true — for the Frobenius algebra to be separable, the complex numbers ${\lambda }_i$ only need to have unit magnitude. They don’t have to be real.

He also said:

Those weightings are precisely the things which give us the TQFT invariants for a genus g surface:

(1)$$Z({\Sigma }_g)={\Sigma }_i{\lambda }_i^{g-1}$$

I think we actually end up summing $\mid {\lambda }_i{\mid }^{2(g-1)}$. After all, the handle operator multiplies the $i$th subspace by $\mid {\lambda }_i{\mid }^2$. We could replace each ${\lambda }_i$ with its magnitude $\mid {\lambda }_i\mid$ and we’d get an isomorphic TQFT, but the TQFTs themselves have this extra degree of freedom. This is why I keep going on about complex numbers rather than just real weightings!

I guess there is not an ArXiv version. I think that was the paper.

My wife is usually my barber. When she neglects that job, I do shave it. But this doesn’t address the fog that is inside my head ;-)

John said:

You say that “every commutative separable Frobenius algebra over the complex numbers looks like ℂ⊕⋯⊕ℂ”. I didn’t know this!

[…] It’s really a fact about algebras, not Frobenius algebras.

I’m really confused now! The result we’re talking about definitely has to do with Frobenius algebras, inasmuch as it’s got the word “Frobenius” in it. I suppose you mean that the proof relies on a fact about plain old algebras — but we’re going to have to connect this fact to Frobenius algebras somehow, which surely won’t be trivial.

Now I’m worried that when you wrote “commutative separable Frobenius algebra”, the “separable” wasn’t being used in its comultiplication-followed-by-multiplication-gives-the-identity meaning! Can you confirm which meaning you meant?

Recall that an algebra is strongly separable iff the bilinear form $$g(a,b)=\mathrm{tr}(L_a{L}_b)$$ is nondegenerate.

OK. Playing with some pictures, I reckon that if a Frobenius algebra has comultiplication followed by multiplication giving the identity, then $\mathrm{tr}(L_a{L}_b)$ will indeed be nondegenerate. It confuses me that you don’t explicitly say this… surely it’s a necessary part of the proof!

I wanted to use ‘special’, the way you do — but then I saw that Schweigert et al use that to mean something slightly different, which agrees in Vect (I think) but not all monoidal categories!

I suppose you’re talking about their definition on page 23. They define “special” to mean that comultiplication followed by multiplication is proportional to the identity — but surely that’s not going to be equivalent to what we’