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 $\not : C\to C$ [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 $C^{op}$, 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 $C^{op}$. 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 $Cat_S$ of categories equipped with this structure, functors weakly but coherently preserving this structure, and natural transformations compatible with this structure.

We know how $Cat_S$ is defined when $S$ is the structure ‘monoidal category’. Is there an automatic way to define $Cat_S$ 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

$$F: Cat_S \to Cat$$

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

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

and an adjoint equivalence between $C$ and $D$, we can lift this adjoint equivalence to one in $Cat_S$.

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 $Iso$ be the walking isomorphism. Then our equivalence

$$1 \simeq Iso$$

automatically gives an equivalence

$$hom(1,C) \simeq hom(Iso,C)$$

for any category $C$. Of course

$$hom(1,C) \simeq C$$

On the other hand, $hom(Iso,C)$ 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 $Equiv$ be the walking equivalence: the 2-category consisting of two objects and an equivalence between them.

We might think that $1$ was equivalent to $Equiv$, 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 $AdEquiv$ be the ‘walking adjoint equivalence’. Then we have

$$1 \simeq AdEquiv$$

so for any 2-category

$$C \simeq hom(1,C) \simeq hom(AdEquiv,C)$$

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

$$C \simeq hom(Iso,C)$$

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 $hom(Iso,C)$ 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 $fg\cong 1_C$ and $gf\cong 1_D$. But there are $\exists$s hiding in the $\cong$s, so we should really add structure. Then we get the data of a map $\alpha:fg\to 1_C$, a map $\beta:gf\to 1_D$, 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

$$e_{ij} = e_i de_j e_j.$$

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 $|\lambda_i| ^{2(g-1)}$. After all, the handle operator multiplies the $i$th subspace by $|\lambda_i| ^2$. We could replace each $\lambda_i$ with its magnitude $|\lambda_i|$ 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 $$g(a,b)=tr(L _a L_ b)$$ is nondegenerate.

OK. Playing with some pictures, I reckon that if a Frobenius algebra has comultiplication followed by multiplication giving the identity, then $tr(L_a L_b)$ 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’re calling special, even in Vect.

They also have this weird extra condition that unit followed by counit equals cup followed by cap. What I don’t like about this is that it seems to rely on the fact that, for each self-dual object, cup followed by cap is well-defined — I don’t know a counterexample, but it would amaze me if this were true in all symmetric monoidal categories. They’re working in modular tensor categories, where you can presumably rely on things like this. Even more confusingly, this condition is redundant, anyway! If cup followed by cap is well-defined, and we have a Frobenius algebra with comultiplication followed by multiplication proportional to the identity, then we’re guaranteed that unit followed by counit is proportional to cup followed by cap, by the same constant of proportionality.

So, I think it’s safe to forget about this extra condition. What we’re calling “special” is what they call “normalised special” — but then they say that they’re just going to call this “special” anyway, because they’re not interested in the non-normalised case.

R. Rosebrugh, N. Sabadini and R.F.C. Walters seem to use use ‘commutative separable algebra’ to mean a commutative Frobenius monoid in any symmetric monoidal category.

Aha — they’re using “commutative separable” to mean “commutative Frobenius, and comultiplication followed by multiplication gives the identity”. On the face of it this seems pretty egregious, seeing that “separable algebra” already means something else. But given what we are talking about above, I reckon it’s true that a commutative algebra which is separable in the traditional sense admits a unique special Frobenius structure, as long we we stick to finite-dimensional complex algebras! So their definition is actually inspired. If this is right, though, then it’s a bit weird that they don’t explain this anywhere.

Maybe I should start by rewriting week268 to call Frobenius algebras with this property … ‘special’ instead of ‘separable’.

Yes, please do this!

We’re talking about the fact that every commutative special Frobenius algebra over the complex numbers is a direct sum of copies of $\mathbb{C}$.

John wrote:

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

Jamie wrote:

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.

Sure — but there’s no real difference between what you call a ‘special Frobenius algebra’ and what algebraists call a ‘strongly separable algebra’. Each one can be made into the other in a canonical way.

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

It’s necesssary, but I figured this part would be easy for you — and it was.

Anyway, let’s be really precise and put an end to any lingering confusion.

A finite-dimensional associative algebra has the property that $tr(L_a L_b)$ is nondegenerate if and only if there exists a way to give it a comultiplication and counit making it into a Frobenius algebra such that comultiplication followed by multiplication is the identity. And, if such a way exists, it’s unique.

So, any special Frobenius algebra (in your sense) has an underlying algebra that is strongly separable. And conversely, any strongly separable algebra can be extended to a special Frobenius algebra in a unique way.

So, classifying special Frobenius algebras is just the same as classifying strongly separable algebras.

The real fun starts here.

John said:

Am I right in assuming that the phrase ‘for which the multiplication is adjoint to the comultiplication’ is a redundant rhetorical flourish, which could safely be omitted?

Yes, you’re right!

Also: if I were defining a †-Frobenius algebra, I’d also want the unit to be adjoint to the counit. Does this follow from the rest somehow?

Normally we would assume that $\dagger$ is strong monoidal, and strong monoidal functors preserve monoids. (Could you even define strong monoidal functors this way, as functors which preserve all monoids? There should be some sort of Strong Microcosm Principle that tells you this would work.) Of course, the $\dagger$-functor is contravariant, so it actually turns monoids into comonoids.

But anyway, all the results I’m stating are just designed to work in FdHilb with its usual $\dagger$-functor and usual tensor product, and this definitely turns monoids into comonoids.

Thanks for answering my first question, Jamie! I don’t think you actually answered the second one. Since I’m trying to get all the details right, I’ll ask you again:

In a $\dagger$-Frobenius monoid, as defined by you and your pals, is the unit the adjoint of the counit?

(I know that $\dagger$ sends monoids to comonoids, but that’s a different issue.)

Could you even define strong monoidal functors this way, as functors which preserve all monoids?

No: there may be very few monoid objects.

Let’s think of a group as a monoidal category with group elements as objects, multiplication as the tensor product, and only identity morphisms. The only monoid object in here is the identity element!

A strong monoidal functor between groups is then exactly the same thing as a group homomorphism. But a functor that preserves monoid objects is just any function sending the identity of one group to the identity of the other.

Wouldn’t we also need the multiplication and unit of our pseudomonoid to be maps, i.e., arise from functors between the source and target categories?

I will need to make some more corrections to handle Robin and Jamie’s objections. I forgot about the requirement for Cauchy completeness, and I guess I’ll need to check to see if anything forces the relevant profunctors to actually be functors — I don’t see any such thing.

I suppose the cup arose from the 1-dimensional world, so gets extended to half a torus, rather than half a sphere.

To really understand all these cups and caps you need to think about 2-manifolds as 2-morphisms in a symmetric monoidal 2-category with duals at all levels.

This counit:

descibes the disappearance of a 1-sphere, or circle. But this process is really the cancellation of an upper semicircle and its dual, the lower semicircle. A counit describes the cancellation of something and its dual.

But the lower semicircle is itself a counit, describing the disappearance of a 0-sphere — that is, the cancellation of the positively oriented point and its dual, the negatively oriented point!

The ‘pairing’ is also a counit:

describing the cancellation of a circle and its dual, an oppositely oriented circle.

But, since the circle is itself the composite of the upper semicircle and the lower semicircle, the pairing can be factored into two separate process, as shown at right. First we do multiplication, which cancels two semicircles:

and then this counit, which cancels two more:

All this stuff is in an old paper that Jim and I wrote.

The whole Frobenius monoid structure of the circle can be derived from this sort of reasoning, and I wish some young and energetic person would do it rigorously.

The unit is profunctor from 1 to $A$, so corresponds to a functor from $A^{op}$ to Set. Looks like it sends $a$ to $Hom(true, not(a))$ or $Hom(a, false)$.

Multiplication is a profunctor from $A \times A$ to $A$, or a functor from $A \times A \times A^{op}$ to Set. Looks like $(a, b, c)$ is sent to $Hom(c, a or b)$.

One unit law relates the sum over $a$ of $Hom(c, a or b) \times Hom(a, false)$ with $Hom(c, b)$.

Hmmm. A pair from the product gives a proof of $b$ from $c$, but are we expecting/achieving isomorphisms? Oh, isn’t there something about degeneracy in classical logic, all proofs of $b$ from $c$ are the ‘same’?

$g(a, b)$ is a composite of multiplication and counit, so of $Hom(c, a or b)$ and $Hom(true, c)$.

David wrote something equivalent to:

??? IS THE FREE SYMMETRIC MONOIDAL BICATEGORY ON A SYMMETRIC FROBENIUS PSEUDOMONOID.

This is a very interesting puzzle!

(Street uses ‘Frobenius pseudomonoid’ instead of ‘categorified Frobenius algebra’; since he invented the definition perhaps we should use his terminology.)

If you take the free symmetric monoidal bicategory on a braided Frobenius pseudomonoid with some extra bells and whistles related to duality, you should get $3Cob_2$ — as Jamie notes, I wrote about this before. I’m not sure Jamie correctly listed all the bells and whistles related to duality; Aaron Lauda has a partially written paper on this conjecture which tackles certain nuances that I’m reluctant to delve into here.

But the question you raise should have an answer that’s not particularly related to 3d topology… since you’re talking about a symmetric rather than braided Frobenius pseudomonoid.

I can’t guess the answer!

John, thanks for explaining this! I am aware that Steve Lack, Ross Street and other Australians are doing amazing stuff in this area but it’s just to far beyond what a half-baked pseudo-categoretician like me can grasp.

Thank you for your quick respons. I was meanly looking for a reason why they introduced frobenius pseudomonoids, soo it’s basically generalizing the connection between frobenius algebras and hopf algebras to Frobenius pseudomonoids and quasi hopf algebras? (correcty me if I’m wrong)

P.s.: Sorry if I’m asking trivial questions, I just wanna be sure.

A Frobenius pseudomonoid is a categorified Frobenius algebra: all the equations in the definition of Frobenius algebra become isomorphisms, and these obey new equations.

So, the reason for introducing Frobenius pseudomonoids is that all the results for Frobenius algebras mentioned in “week268”—and others, too—deserve to be categorified.

This is nicely explained in Aaron Lauda’s two papers.

so it’s basically generalizing the connection between frobenius algebras and Hopf algebras to Frobenius pseudomonoids and quasi Hopf algebras?

Quasi Hopf algebras are not categorified Hopf algebras, so no, this doesn’t sound right.