### Star-Structures and Daggers

#### Posted by Urs Schreiber

*A question by Bruce Bartlett:*

Hi guys,

I’ve got a question about duality for 2-categories. John Baez and Laurel Langford defined what a “monoidal 2-category with duals” was in HDA IV. The basic concept is easy enough to understand. A monoidal 2-category with duals is a 2-category with duals on all levels : duals for objects, morphisms and 2-morphisms.

Thus every 2-morphism $\theta : F \Rightarrow G$ has a dual $\theta^* : G \Rightarrow F$, every morphism $F : A \rightarrow B$ has a dual $F^* : B \rightarrow A$ and every object $A$ has a dual $A^*$. That’s the basic picture.

For our purposes here, we can ignore the tensor product and the duals for objects side of things, so don’t worry about that.

Since this is a long post, for the experts I’ll state my question right up. Can anyone help me understand the equation $(\theta^\dagger)^* = (\theta^*)^\dagger ?$

.

Ok, lets explain this. Imagine we have a 2-category where every 2-morphism $\theta : F \Rightarrow G$ has a dual $\theta^* : G \Rightarrow F$, satisfying the obvious axioms like $(\theta^*)^* = \theta$ and compatibility with vertical and horizontal composition: $(\theta \circ \phi)^* = \phi^* \circ \theta^* \quad , \quad (\theta * \phi)^* = \theta^* * \phi^*$ where $\circ$ and $*$ denote vertical and horizontal composition respectively.

You want to be thinking of examples like “2-tangles” at this point, or “2-Hilbert spaces”. Or even “hermitian coherent sheaves” - if there are such a thing.

Ok, lets draw all of this in string diagrams [don’t try upload this post at home!!]. So for every 2-morphism $\theta : F \Rightarrow G$,

,

we have a dual 2-morphism $\theta^* : G \Rightarrow F$,

.

The arrows are there for our next step. Suppose that every morphism $F : A \rightarrow B$ in our category has a left adjoint $F^* : B \rightarrow A$. As usual, we draw these in string diagrams by giving orientations to the edges:

The unit $\eta : \id_A \Rightarrow F^*F$ and counit $\epsilon : F F^* \Rightarrow \id_B$ maps are drawn as

and they satisfy the snake diagrams

.

If $\theta : F \Rightarrow G$ is a 2-morphism, we can use the left adjoints for $F$ and $G$ to make a 2-morphism $\theta^\dagger : G^* \Rightarrow F^*$ [Ed : some people call * this * the “star”]:

.

Ok. Standard stuff so far. Now, the $*$-structure on the 2-morphisms allows us to make $F^*$ also into a * right * adjoint of $F$, with unit $\epsilon^*$ and counit $\eta^*$.

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 $(\theta^\dagger)^* = (\theta^*)^\dagger.$ In string diagrams,

.

In the case of 2-tangles, John and Laurel proved that these define the same 2-morphism. In the case of 2Hilb, I am unable to prove it elegantly except by brute force calculation! One has to choose a basis (2-Hilbert spaces are semisimple categories), etc.

Can anyone (John?) explain the significance of this equation, $(\theta^\dagger)^* = (\theta^*)^\dagger$? In the context of coherent sheaves, it seems that this equation has to do with Serre duality (though I don’t understand this stuff). More precisely, Serre duality seems
somehow to have to do with the * failure * of this equation to hold
- one needs to “twist” the left or right hand side appropriately. - one has to twist the left or right hand sides first.

Anyhow, when one works with 2-characters of 2-representations, it seems that one needs this equation to hold, eg. in 2Hilb. I’d like to understand it more deeply!

## Re: Star-Structures and Daggers

Bruce wrote:

You almost make it sound like we did something nontrivial here. In fact, it’s completely obvious from the picture you just drew! Just take the equation, stretch out both sides, and they straighten out to become the same thing. QED.

This sort of ‘picture proof’ isn’t rigorous in every context, but it is for 2-tangles. Since we were just trying to abstract the algebraic structure from 2-tangles and make it into definitions about duality for monoidal 2-categories, we were completely content.

Speaking of pictures, you have a $\theta$ on the right side of this equation:

where I would write either $\theta^*$ or a $\theta$ reflected across the $x$ axis. I find that confusing. I guess you’re making the labelling of strings do the work in telling us when we’ve got a $\theta$ and when we’ve got a $\theta^*$. I’d probably get mixed up if I tried to do that.

In fact, I would use a less symmetrical letter than $\theta$ for your 2-morphism here. How about $\rho$? Then in terms of pictures, $\rho^*$ is $\rho$ reflected across the $x$ axis, while $\rho^\dagger$ is $\rho$ rotated 180 degrees — at least after you straighten out those strings. Since a reflection commutes with a 180 degree rotation, $(\rho^\dagger)^* = (\rho^*)^\dagger$. Both of these are just $\rho$ reflected across the $y$ axis!

But it sounds like you want another, more

algebraicunderstanding of the equation$(\rho^\dagger)^* = (\rho^*)^\dagger.$

I have a feeling Todd Trimble could tell you some interesting things about that. But here are a few remarks.

Before you seek a simple, elegant proof of this fact in $2\Hilb$ — which surely must exist! — it’s good to find a simple proof that

$(\rho^\dagger)^* = (\rho^*)^\dagger$

for any 1-morphism

$\rho : F \to G$

in $\Hilb$ — that is, any linear operator $\rho$ from a finite-dimensional Hilbert space $H$ to a finite-dimensional Hilbert space $G$.

Here

$\rho^\dagger: G^* \to F^*$

is the usual adjoint of linear operators between vector spaces:

$(\rho^\dagger \ell)(v) = \ell (\rho v)$

for any linear functional $\ell \in G^*$ and vector $v \in F$. On the other hand,

$\rho^*: G \to F$

is defined using the inner product:

$\langle \rho^* w, v \rangle = \langle w, \rho v \rangle$

for all $v \in F$, $w \in G$.

Surely there’s some easy, basis-free proof that

$(\rho^\dagger)^* = (\rho^*)^\dagger.$

in this context. Can you boost it up a dimension and do something similar in $2Hilb$?