Calculations Inside Semisimple Categories

May 18, 2007

Calculations Inside Semisimple Categories

Posted by Urs Schreiber

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– guest post by Bruce Bartlett –

Hi guys,

I’ve got a question regarding performing calculations inside semisimple categories.

I posted a version of it back in March. I’ve made quite a lot of progress, but I’m still missing some vital ingredient which I can’t put my fingers on. John hinted (rather sneakily!) in one of his TWF comments that he had made some progress on it too. Since everyone has been working so diligently on it, I thought it’s time to reconvene the homework group and compare notes.

Recall that the problem is about defining adjoints (or ‘daggers’) of natural transformations. There are two ways to define them : a left-handed way and a right-handed way, and we’d like to check they are the same.

Don’t worry about the specific problem. Just think : we have some calculation to perform inside a semisimple category. What are we going to do?

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Since we want to be as elegant as possible, here are the ground rules :

(a) No assuming that the categories are skeletal.

(b) Choosing bases of the hom-spaces is to be frowned upon.

Ok. Now let me explain the progress I’ve made, and then perhaps people can comment on how they would go about performing this type of calculation. The bad news first : my method hasn’t yet been able to solve the problem! $pic$

We perform calculations the only way I know how - by developing a graphical calculus for working with semisimple categories enriched in Hilb (or Vect). In fact, it’s just a simplified version of John’s QG notes on a graphical calculus for working with closed monoidal categories.

Right, let’s begin. We have a semisimple category $H$ , which I’m going to sometimes assume is a 2-Hilbert space. That basically just means it’s enriched in Hilb; i.e. there are inner products on the hom-spaces.

Anyhow, for any two objects $x,y \in H$ we have a vector space $hom(x,y) \in Vect$ , which I’ll occasionally write simply as $(x,y)$ , and we draw it as a ribbon:

I’m too lazy to colour these ribbons in… sue me! Composition of morphisms in $H$ corresponds to a multiplication map $m : hom(x,y) \otimes hom(y,z) \rightarrow hom(x,z)$ :

Remember, these diagrams are happening inside the monoidal category Vect, and they go top-down. The identity morphism $id_x : x \rightarrow x$ for each object $x$ corresponds to a map $1_x : \mathbb{C} \rightarrow hom(x,x)$ :

Right, so far everything has been standard. Now let’s specialize to 2-Hilbert spaces. Since they’re enriched in Hilb, we can take the duals (adjoints) of all these linear maps inside Hilb. So we have the adjoint of the multiplication map, $m^* : hom(x,z) \rightarrow hom(x,y) \otimes hom(y,z)$ and also the adjoint of the identity, $1_x^* : hom(x,x) \rightarrow \mathbb{C}$ :

We haven’t used semisimplicity yet. Being semisimple means there are a finite number of nonisomorphic simple objects $e_i \in H$ . It’s well known (see page 12 of John’s 2-Hilbert spaces paper) that this implies we can decompose any morphism $f : x \rightarrow y$ canonically into its ‘isotypic’ components,

(1)

f = \Sigma_i f_i.

I realized recently that a nice way to express this is to say that we have a canonical ‘resolution of the identity’:

(2)

hom(x,y) \cong \oplus_i hom(x, e_i) \otimes (e_i, y)

This makes it clear that it’s the categorification of the corresponding statement for ordinary Hilbert spaces,

(3)

(x,y) = \Sigma_i (x, e_i) (e_i, y).

Let’s be clear about the equation occuring two lines up. When I use the objects “ $e_i$ ” above, I haven’t chosen a basis on the hom-sets. It works for any complete set of simple objects, and its canonical. Going from right-to-left, one sends a pair $f : x \rightarrow e_i$ , $g : e_i \rightarrow y$ to

(4)

f \otimes g \mapsto g \circ f

That’s canonical, so its inverse is too. It is analogous to the statement that for any vector spaces $V,W$ , there is a canonical isomorphism $Hom(V,W) \cong V^* \otimes W$ .

Anyhow, we draw the resolution of the identity $hom(x,y) \cong \oplus_i hom(x,e_i) \otimes hom(e_i, y)$ as

It’s inverse is the multiplication map, so we have the rules

You get the idea.

And so we can go on and on. We also have the symmetry map

$hom(x,y) \otimes hom(y,z) \rightarrow hom(y,z) \otimes hom(x,y),$

as well as the ‘star’ map

$* : hom(x,y) \rightarrow hom(y,x),$

It’s pretty cool how the $*$ -structure makes our ribbons behave seriously as ribbons! And you can do functors, natural transformations, adjoints, etc. all in this graphical language. Here at the n-category cafe, I’m preaching to the choir.

This is getting a bit lengthy… so I’ll just quickly recall the calculation we need to perform, which I was hoping could be done in a nice graphical way.

Problem : We have linear direct sum-preserving functors $F, G : H \rightarrow H^'$ , which have adjoints (semisimple categories!) $F^* , G^* : H^' \rightarrow H$ , and we have a natural transformation $\theta : F \Rightarrow G$ .

We want to concoct the corresponding natural transformation between their adjoints, $\theta^\dagger : G^* \Rightarrow F^*$ .

Remarks. Sit down for a second and you’ll see that there are two natural choices, which I drew in string diagrams in my last post. Since $F^*$ is the (left-and-right) adjoint for $F$ , we have isomorphisms

(5)

\phi_F : (F x, y) \stackrel{\cong}{\rightarrow} (x, F^*y), \quad \psi_F : (y, F x) \stackrel{\cong}{\rightarrow} (F^*y, x)

and the same for $G$ . Of course, $\phi$ and $\psi$ are just ‘flip sides of the same coin’, i.e.

(6)

\phi = * \psi *,

or in pictures:

Anyway, we have to calculate if the following two candidates $\theta^\dagger_L, \theta^\dagger_R : G^* \Rightarrow F^*$ are the same. In other words, we need to check whether the following two elements of $(G^*y, F^*y)$ are the same. $(\theta^\dagger_L)_y$ is defined as:

(7)

id \in (G^*y, G^*y) \stackrel{\phi_G^{-1}}{\rightarrow} (G G^*y, y) \stackrel{pre(\theta)}{\rightarrow} (F G^*y, y) \stackrel{\phi_F}{\rightarrow} (G^*y, F^*y)

While $(\theta^\dagger_R)_y$ is defined as:

(8)

id \in (F^*y, F^*y) \stackrel{\psi_F^{-1}}{\rightarrow} (y, F F^*y) \stackrel{post(\theta)}{\rightarrow} (y, G F^*y) \stackrel{\psi_G}{\rightarrow} (G^*y, F^*y)

Now, are these the same? I’ve tried lots of graphical calculations… there’s even another way to define $\theta^\dagger$ … but I’m sure I’m missing something, perhaps something very simple. Can anyone help?

Posted at May 18, 2007 12:22 PM UTC

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Re: Calculations Inside Semisimple Categories

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I’m sorry not to have replied sooner… if anyone should answer this, it’s me!

It’s possible your long, nice, pedagogical description of the problem made it seem more specific, technical, and terrifying than it really is. There might be some way to strip away the details of $2Hilb$ and ask this question for a bunch of 2-categories. If that’s true, folks like Todd Trimble might be able to solve it in a jiffy.

(You might object that there’s nobody ‘like’ Todd Trimble. However, I know at least one.)

Here’s an attempt. Let me know if an answer to this question would make you happier.

Suppose we have a couple of objects $x$ and $y$ in a 2-category $C$ . Suppose we have morphisms

$F, G : x \to y$

and a 2-morphism

$\theta : F \Rightarrow G$

Suppose $F$ has a left adjoint $F^*$ , and $G$ has a right adjoint $G^*$ . Then, I think there’s just one way to define a 2-morphism

$\theta^\dagger : G^* \Rightarrow F^*$

It looks like this:

         |
         |
         |
       G*|              _______
         |            F/       \
         |            /         \
         |           θ           |
          \         /            |
           \_______/G            |
                                 |F*
                                 |
                                 |
                                 |

(Pardon the low-tech ASCII art — I just happened to have this figure lying around.)

But now, suppose $F^*$ is also the right adjoint of $F$ , and $G^*$ is also the left adjoint of $G$ . This gives a second way to define a 2-morphism from $G^*$ to $F^*$ :

                                 |
                                 |
                                 |
            _______              |
           /       \F            |G*
          /         \            |
         |           θ           |
         |            \         /
       F*|            G\_______/
         |
         |
         |
         |

Does this have to be the same?

If so, why?

Personally, I can’t imagine a general abstract-nonsense argument that these two 2-morphisms, say $\theta^\dagger_L$ and $\theta^\dagger_R$ , need to be the same… not at this level of generality, anyway!

If no such argument exists, maybe this is some extra coherence law we can place on a 2-category where every morphism is part of an ambidextrous adjunction. And then you’re either asking whether this law holds in $2Hilb$ , or asking for a diagrammatic calculation that proves this law. Which one is it?

Can’t you just do a grungy basis-ridden linear algebra calculation to see whether this law is true in $2Hilb$ ? If it is, then there must be some good reason. At worst, you can just add this law to your bag of string diagram calculation techniques.

If it’s not true, I’ll be shocked! I’ll buy you a pizza.

Posted by: John Baez on June 2, 2007 3:59 AM | Permalink | Reply to this

Re: Calculations Inside Semisimple Categories

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John wrote:

…maybe this is some extra coherence law we can place on a 2-category where every morphism is part of an ambidextrous adjunction.

Hmm. Isn’t a monoidal category with both left and right duals called “pivotal” if this coherence law holds? I seem to remember that from the work of Yetter. I think Barrett and Westbury mention this work in their paper on “spherical” monoidal categories.

If there’s a name for this law, it must mean this law isn’t automatically true…

Posted by: John Baez on June 2, 2007 4:17 AM | Permalink | Reply to this

Re: Calculations Inside Semisimple Categories

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Hooray, someone answered my post :-) Good news! A couple of days ago I figured out how to answer this question… and greatly improved my calculational prowess inside semisimple categories :-)

The “left-handed” and “right-handed” alternatives for defining the adjoint of the natural transformation do turn out to be equal.

But here’s the bad news: I don’t think I have a deeper understanding of precisely why they are equal. As you suggested, I will look into this phenomenon one level down, where it features in the definition of “pivotal categories”. All I can say is, to prove they’re equal, I’ve had to use the $*$ -structure and inner products on 2Hilb quite a lot.

I’d like to remind everyone at this point that a nice place to see duality playing itself out nontrivially (via Serre functors, etc.) in a 2-category context is in the 2-category Var of complex manifolds and integral kernels, as I pointed out in this post.

Okay, here’s the way I now like to work with semisimple categories like 2-Hilbert spaces, and how it helped me to solve the problem :

One should think of [semisimple categories, functors between them, and transformations between those] via the hom-sets.

That’s similar to the “physicist’s” way, eg. Moore and Seiberg’s original conformal field theory paper, or chapter 5.3 of Bakalov and Kirillov’s book.

Basically we have the correspondence:

(1)

2-Hilbert space H \quad \leftrightarrow \quad finite collection X of points

(2)

functor F: H \rightarrow H^' \quad \leftrightarrow \quad vector bundle over H^' \times H

(3)

natural transformation \theta : F \Rightarrow G \quad \leftrightarrow \quad morphism of vector bundles

I know that’s practically the same as the bad-old-way of thinking of 2-vector spaces as having “objects the natural numbers, morphisms are matrices of vector spaces, and 2-morphisms are linear maps between the matrices”. It is in it’s final result … but now how you got there . The latter way usually involves a kind of choice of a skeleton, while the former doesn’t.

Suppose we have two 2-Hilbert spaces $H$ and $H^'$ . I ask you to pick a maximal set of nonisomorphic simple objects $e_i$ and $e^'_\mu$ for $H$ and $H^'$ . That’s a finite choice… it’s not like choosing a skeleton! And it’s the last choice you’ll ever have to make. Human beings are finite creatures.

Thus to $H$ we have assoicated the index set $X$ (the “i’s” of the simple objects $e_i$ ), and to $H^'$ we have the index set $X^'$ .

To a linear functor $F : H \rightarrow H^'$ we associate the vector bundle over $X^' \times X$ whose fiber at $(\mu, i)$ is the “categorified transition probability for going from $i$ to $\mu$ ”:

(4)

F_{\mu, i} = \langle \mu |\, F\, | i \rangle := Hom(e^'_\mu, F(e_i)).

So… the point is that the quantum mechanical “tao” is as useful to 2-Hilbert spaces as it was for Hilbert spaces! Now the chance of going from $i$ to $\mu$ is a vector space, not a number! In fact, it was when I tried to reply to John’s post about this topic that I realised I should push this quantum mechanical analogy for all it’s worth (I’m sure John will be pleased!), as I’ll show you soon.

A natural transformation $\phi : F \Rightarrow G$ get’s sent to a morphism of vector bundles, whose components

(5)

\phi_{\mu, i} : \langle \mu |\, F\, | i \rangle \rightarrow \langle \mu |\, G\, | i \rangle

are given by post-composition,

(6)

a \mapsto post(\phi_{e_i}) \circ a.

Up to now everything would work for any semisimple category. A nice property of 2-Hilbert spaces is that a natural transformation $\phi : F \Rightarrow G$ is naturally, freely and uniquely determined (not just “up to isomorphism… bah!”) by it’s corresponding map of vector bundles, as perhaps we’ll see soon. This is a stronger statement than saying that “every natural transformation is freely and uniquely determined by it’s components on the simple objects $e_i$ ”.

But first I want to point out the usefulness of using the Dirac notation,

(7)

\langle i | F | \mu \rangle = Hom(e^'_\mu, F(e_i)).

If we have another functor $G : H^' \rightarrow H^{' '}$ , then this notation works as for Hilbert spaces. Namely there is a canonical isomorphism

(8)

\langle y | G F | i \rangle \cong \oplus_\mu \langle y | G | \mu \rangle \otimes \langle \mu | F | i \rangle

given by sending a pair $f : e^'_u \rightarrow F(e_i)$ , $g : e^{' '}_y \rightarrow G(e^'_\mu)$ to $G(f) \circ g$ .

This post has become too long, so I guess I’ll refrain from actually showing how this notation helped me to perform the calculation. If anyone is interested, just ask.

I can’t help mentioning a nice way to handle adjoints using the bra-ket notation. Namely, we should really be adding “double-bars” and writing

(9)

\langle \mu || F | i \rangle := Hom(e^'_\mu, F(e_i)).

The double bars are “preventing $F$ from acting on the left”… unfortunately MathML overdoes them a bit. This enables us to incorporate (right) adjoints by writing

(10)

\langle \mu | F || i \rangle := Hom(F^*(e^'_\mu), e_i).

The adjoint isomorphism then becomes an isomorphism

(11)

\psi : \langle \mu || F | i \rangle \stackrel{\cong}{\rightarrow} \langle \mu | F || i \rangle

And so on… there are a few more tricks to play.

Posted by: Bruce Bartlett on June 2, 2007 11:27 AM | Permalink | Reply to this

Re: Calculations Inside Semisimple Categories

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Yay! I’m glad you figured out. I’d like to see a bit more about how you did it. I’m quite happy with your categorified bra-ket notation; it would be fun to see how it gets the job done. Then maybe we can find a deeper understanding of why it all works.

I have my own reasons for wanting to understand this stuff, now. For one, my student Jeffrey Morton is finishing up his thesis, and at the last minute we realized that some tricky 2-Hilbert space computations are required… which so far are much less elegant than we’d like! It might be a related issue at work. I’m not sure, yet.

I just noticed that the pivotal law, namely:

         |
         |
         |
       G*|              _______
         |            F/       \
         |            /         \
         |           θ           |
          \         /            |
           \_______/G            |
                                 |F*
                                 |
                                 |
                                 |
                 = 
                                 |
                                 |
                                 |
            _______              |
           /       \F            |G*
          /         \            |
         |           θ           |
         |            \         /
       F*|            G\_______/
         |
         |
         |
         |

doesn’t explicitly mention duals for 2-morphisms. It merely requires that we have ambidextrous adjoints for 1-morphisms.

But, in $2Hilb$ , it’s the duals for 2-morphisms that make the adjoints of 1-morphisms be ambidextrous! And, you hint that your proof of the pivotal axiom in $2Hilb$ really uses the duals for 2-morphisms.

So, it seems that the pivotal axiom is all about the interaction between duality at two levels — the 1-morphism level and the 2-morphism level.

Does this seem right to you?

By the way: in your comment, you seem to be spending a lot of energy on moral scruples of various sorts:

I know that’s practically the same as the bad-old-way of thinking of 2-vector spaces as….

That’s a finite choice… it’s not like choosing a skeleton! And it’s the last choice you’ll ever have to make.

not just “up to isomorphism… bah!”

It’s dangerous to get too fussy about how one solves a problem before one has solved it. Especially if your moral scruples are self-contradictory! Not wanting to pick a skeleton means you’re avoiding equations when isomorphisms will do — I’m all in favor of that. But complaining about things that are only defined up to isomorphism reflects an exactly opposite moral stance: now you’re insisting on having things defined up to equality, when isomorphism will do. Having an inconsistent moral code can make life rather tough.

In math, moral scruples are great when they help you solve problems. They’re not so great when they hem you in so much that it gets harder to solve problems.

That’s why I speak of the ‘tao of mathematics’. The tao is not a set of moral scruples: tao is Chinese for ‘way’, and the idea is to do everything the way it’s most easily done, just like water runs downhill in a river bed.

Posted by: John Baez on June 2, 2007 4:55 PM | Permalink | Reply to this

Re: Calculations Inside Semisimple Categories

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Hi John, thanks for your comments. I agree with all that you say, after all, I learnt all of this stuff from you!

So, it seems that the pivotal axiom is all about the interaction between duality at two levels; the 1-morphism level and the 2-morphism level. Does this seem right to you?

Yes, I would agree with that. After all, that’s precisely the way you put it in your 2-tangles paper! In axiom 12 of HDA IV , axiom 12 of a “monoidal 2-category with duals” requires that

(1)

For any \text{ 2-morphism} \alpha, {\alpha^\dagger}^* = (\alpha^*)^\dagger.

And that’s precisely the equation we are trying to prove. I spoke about this back in this post.

An interesting point though, in this regard, is that it turns out that the natural transformation

(2)

\theta^\dagger : G^* \Rightarrow F^*

(defined either in the left or the right way, since they are the same), is actually independent of the actual choice of the adjoint structures on $G^*$ and $F^*$ !

By “adjoint structures”, I mean it’s independent of choice of the co-unit and unit for $F^*$ and $G^*$ . Obviously it’s not independent of the choice of the actual functors $F^*$ and $G^*$ … that wouldn’t make sense.

I find this very surprising, and I’m still trying to get to the bottom of it. In the end everything does actually look similar to the proof for ordinary Hilbert spaces, despite my doubts to the contrary as I expressed in this post.

In that proof, one proves that the LHS = RHS because they’re both equal to the “Middle Hand Side” - which is the line

(3)

= (\rho(a), b) by [1]

in that proof.

And a similar thing happens for 2-Hilbert spaces.

Regarding your comments about my moral scruples; sorry about my silly remarks. The goal of the project Simon has outlined for me is to use this “2-representations, higher categories and TQFT technology” as a tool to study the geometry of eg. Chern-Simons theory. I guess I’ll never actually get that far in the end…

But when you’re trying to extract geometrical information from categories, you have to be very careful to do everything as canonically as possible. Geometry is all about trying to be as canonical as possible… otherwise you couldn’t tell the difference between a vector bundle and a trivial vector bundle. Both of them are just fiber bundles over a base space whose fibers are isomorphic to $\mathbb{C}$ .

And that’s the sort-of conundrum a geometrically minded person faces when he tries to use higher categories. There is a trade-off between brute force involving lots of choices and computational power on the one hand, and less choices, more canonically defined structures… but the information is harder to extract.

We know that’s the case for associators for monoidal categories. There’s all sorts of interesting geometrical data there… but because these categories are so big, this data is kind of hidden. You can use brute force to get it out, but then you sometimes do violence to the geometry, and it’s a catch-22.

So I’m constantly in a state of tension between these two extremes, and I occasionally make silly remarks!

Posted by: Bruce Bartlett on June 2, 2007 6:42 PM | Permalink | Reply to this

Re: Calculations Inside Semisimple Categories

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John wrote:

Now that you point it out, I remember that you mentioned the law $(\alpha^*)^\dagger = (\alpha^\dagger)^*$ aeons ago. But were you talking about the pivotal law back then, too?

Indeed :-) Halfway down that post, I defined $\alpha^\dagger$ in terms of the left adjoints, and then after explaining that the $*$ -structure allows us to turn a left adjoint into a right adjoint, I made the comment,

This means we could also have used the right adjoints to define the daggers. Do we get the same answer? If you draw these out, you’ll see that’s the same question as asking whether $(\alpha^\dagger)^* = (\alpha^*)^\dagger$ .

To see this, just take $*$ ’s of the 2-morphisms on both sides. In other words,

(1)

(\alpha^\dagger)^* = (\alpha^*)^\dagger \quad \Leftrightarrow \quad \alpha^\dagger = ((\alpha^*)^\dagger)^*.

And that latter equation is precisely the pivotal law! Since if you write it out, you’ll see that

(2)

\alpha^\dagger = \alpha_L^\dagger \quad and \quad (\alpha^*)^\dagger)^* = \alpha_R^\dagger.

This is just a matter of using the rules $(\alpha \beta)^* = \beta^* \alpha^*$ and $(\alpha \box \beta)^* = \alpha^* \box \beta^*$ , where $\box$ denotes horizontal composition (MathML doesn’t recognize \ast), and the fact that we’ve chosen the co-units for the right adjunctions to be the $*$ ’s of the units for the left adjunctions, etc.

I originally chose to express the equation in the form

(3)

\alpha_L^\dagger \stackrel{?}{=} \alpha_R^\dagger

because that makes no explicit use of the $*$ -structure; this question can be asked whenever you have an ambidextrous adjunction. And I thought that would give it a wider appeal - I know that many people aren’t familiar with “ $*$ -structures” and “2-Hilbert spaces” so I wanted to express it in a language everyone could understand.

Now with regard to your second point, about my statement that the value of $\alpha^\dagger$ is independent of the choice of adjoint structures on $F^*$ and $G^*$ . You wrote,

This is a well-known result. It applies to any adjunction in any bicategory. … and then you’ll see that, up to these comparison isomorphisms, the two ways of defining $\alpha^\dagger$ are the same.

I think you misunderstood me. I was claiming that $\alpha^\dagger$ is truly independent of the choice of co-units and units for $F^*$ and $G^*$ … not just “up to a comparison isomorphism”. Indeed I’m aware of the canonical isomorphism between two different adjoints; one just draws out the obvious morphism in string diagrams. I learnt this ages ago from the Fall 2004 QG seminar (Derek Wise’s homework answer to be precise!)

But I was wrong in my claim anyhow, so it doesn’t matter! I’ve rechecked the calculation; $\alpha^\dagger$ does depend on the choice of adjoint structures… it just depends covariantly, in the obvious way, as you were describing. Sorry about that; we’ve both come to the same conclusion in the end.

Posted by: Bruce Bartlett on June 3, 2007 11:12 AM | Permalink | Reply to this

Re: Calculations Inside Semisimple Categories

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I was so annoyed by my failure to notice — or remember — that

${\alpha^*}^\dagger = {\alpha^\dagger}^*$

is none other than the pivotal law

         |
         |
         |
       G*|              _______
         |            F/       \
         |            /         \
         |           α           |
          \         /            |
           \_______/G            |
                                 |F*
                                 |
                                 |
                                 |


                 = 


                                 |
                                 |
                                 |
            _______              |
           /       \F            |G*
          /         \            |
         |           α           |
         |            \         /
       F*|            G\_______/
         |
         |
         |
         |

that when I finally noticed this (while making dinner), I ran to the computer and deleted the comment you just replied to. However, due to a bug in the blogging software, the deletion didn’t kick in until much later — after you saw the comment and replied.

Oh well, no big deal. I’m mentioning this just in case anyone is wondering how you replied to a nonexistent comment!

I guess the main upshot of all this is, as you say, that while the ‘pivotal’ condition makes sense in any 2-category where all 1-morphisms have ambidextrous adjoints, it’s quite mysterious unless we assume 2-morphisms have duals — at which point it becomes a quite natural condition, though not automatic.

Posted by: John Baez on June 3, 2007 4:39 PM | Permalink | Reply to this

Re: Calculations Inside Semisimple Categories

This was a very amusing episode. I have the dubious distinction of having replied to a non-existent comment! Reminds me of the Great Existence Debate.

Thanks for your replies, John. It’s helped me to put my thoughts in order.

Posted by: Bruce Bartlett on June 3, 2007 6:13 PM | Permalink | Reply to this

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May 18, 2007