### Calculations Inside Semisimple Categories

#### Posted by Urs Schreiber

*– 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?

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!

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,

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

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

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

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

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

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:

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

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?

## Re: Calculations Inside Semisimple Categories

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

leftadjoint $F^*$, and $G$ has arightadjoint $G^*$. Then, I think there’s just one way to define a 2-morphism$\theta^\dagger : G^* \Rightarrow F^*$

It looks like this:

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

But now, suppose $F^*$ is also the

Does this have to be the same?rightadjoint of $F$, and $G^*$ is also theleftadjoint of $G$. This gives a second way to define a 2-morphism from $G^*$ to $F^*$: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

whetherthis law holds in $2Hilb$, or asking for adiagrammatic calculationthat proves this law. Which one is it?Can’t you just do a grungy basis-ridden linear algebra calculation to see

whetherthis 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.