## March 15, 2007

### 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!

Posted at March 15, 2007 9:08 PM UTC

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### Re: Star-Structures and Daggers

Bruce wrote:

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.

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 algebraic understanding 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 $2Hilb$?

Posted by: John Baez on March 16, 2007 7:19 AM | Permalink | Reply to this

### Re: Star-Structures and Daggers

It’s also related to the fact that theta commutes with the balancing of the category in question.

To see this, start on the left. Bend G up to the right and F down to the left. We can straighten that out by using the “snake diagrams”. On the left, though, we can pull the strands past theta (naturality) to get a positive writhe on F and a negative writhe on G. Now add a positive writhe on G to both sides of the equation: we have theta composed with the balancing on G on one side, and composed with the balancing on F on the other.

I hope my description is clear…

Posted by: John Armstrong on March 16, 2007 12:48 PM | Permalink | Reply to this

### Re: Star-Structures and Daggers

Kind of clear… I think . I won’t put you through the torture of having to actually draw those diagrams inside this iTeX environment!

But, aren’t you thinking in the context of “balancings” inside a braided monoidal category? I’m talking about a 2-category with duals at all levels. The equation $(\theta^\dagger)^* = (\theta^*)^\dagger$ is an equation concerning 2-morphisms inside a 2-category, not morphisms inside a category.

On the other hand, these two contexts are clearly related : after all a monoidal category with duals (at all levels) is basically a 2-category with duals (at all levels), with only one object. To put the braiding into this picture, we need to start talking about a 3-category with duals (at all levels), with only one object. That should be the same as a braided monoidal category with duals. But lets not go there… unless you feel like dipping into Nick Gurski’s thesis :-)

Posted by: Bruce Bartlett on March 16, 2007 3:03 PM | Permalink | Reply to this

### Re: Star-Structures and Daggers

Okay, I left that bit out. Basically a balancing is an attempt to make “twists” in a ribbon category. When you’re drawing string diagrams you can get the same thing by drawing a writhe.

I might be wrong about the details, and please tell me if this doesn’t check out. Start from the beginning and draw theta in a box with F going up and G going down. Add a curl in the line describing F. Now zig-zag the top and bottom lines, one in each direction. Slide around the lines by naturality and using your equation. You should be able to end up with theta in a box and the curl on the line describing G below theta.

That shows that your equation implies theta commutes with the “curl”. You should be able to do it the other way as well to get the opposite implication.

Posted by: John Armstrong on March 16, 2007 11:09 PM | Permalink | Reply to this

### Re: Star-Structures and Daggers

Doesn’t Hilb have an isomorphism $A\cong A^*$ under which $(-)^*$ and $(-)^\dagger$ become identified? In fact, it seems that you need this isomorphism in order to define what you’ve called $\rho^*$. So the equation $(\rho^*)^\dagger = (\rho^\dagger)^*$ becomes trivial, since both of them are isomorphic to $\rho$?

Posted by: Mike Shulman on March 16, 2007 10:44 PM | Permalink | Reply to this

### Re: Star-Structures and Daggers

Mike writes:

Doesn’t Hilb have an isomorphism $A\cong A^*$ under which $(-)^*$ and $(-)^\dagger$ become identified?

Not if $Hilb$ is the category of complex Hilbert spaces, which is the kind physicists like. For a complex Hilbert space $H$, the map

$\begin{matrix} H &\to& H^* \\ v &\mapsto & \langle v, \cdot \rangle \end{matrix}$

is 1-1 and onto, continuous, with continuous inverse — but conjugate-linear, so it’s not a morphism in $Hilb$!

(I’m following physicists and making my inner product linear in the second slot. Fiddling with this sort of thing can’t save us.)

So, one really needs to keep track separately of the vector space adjoint $F^\dagger : K^* \to H^*$ and the Hilbert space adjoint $F^*: K \to H$ of a bounded linear operator $F: H \to K$.

Another approach, ultimately equivalent, is to keep track of dual vector spaces and the way the Galois group of $\mathbb{C}$ acts on things. For every complex Hilbert space $H$, there’s a complex conjugate Hilbert space $\overline{H}$. The above map gives a natural isomorphism $\overline{H} \cong H^*.$

I spent about 5 years sifting my way through this stuff… let’s not even talk about quaternionic Hilbert spaces.

Posted by: John Baez on March 17, 2007 3:02 AM | Permalink | Reply to this

### Re: Star-Structures and Daggers

John wrote:

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

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

in this context.

Okay, let me have a hand at this.

We have a linear map $\rho : A \rightarrow B$ between finite dimensional Hilbert spaces, and we form the adjoint $\rho^* : B \rightarrow A$ and “dual” $\rho^\dagger : B^* \rightarrow A^*$ (see John’s post for the notation).

We’ll use five facts:

1. The adjoint map $\rho^* : B \rightarrow A$ satisfies $(\rho(a), b)_B = (a, \rho^*(b))_A$.

2. Taking complex conjugates gives $(b, \rho(a))_B = (\rho^*(b), a))_A$.

3. Every linear functional on a Hilbert space $A$ is uniquely of the form $(a, \cdot)$ for some vector $a \in A$.

4. This gives an inner product on the dual space $A^*$ by $((a_1, \cdot)_A, (a_2, \cdot )_A )_{A^*} = (a_1, a_2)_A$.

5. If $\rho : A \rightarrow B$, then the dual map $\rho^\dagger : B^* \rightarrow A^*$ computes as $(a, \cdot)_B \mapsto (a, \rho(\cdot))_B$.

Lemma : If $\rho : A \rightarrow B$, then $(\rho^\dagger)^* = (\rho^*)^\dagger$.

Proof : Their inner products are the same, hence they must be the same map:

$\left( (\rho^\dagger)^* (a, \cdot), (b, \cdot) \right) = \left((a, \cdot), \rho^\dagger(b, \cdot) \right) \quad by$ 

$= \left( (a, \cdot), (b, \rho(\cdot)) \right) \quad by$ 

$= \left( (a, \cdot), (\rho^*(b), \cdot) \right) \quad by$ 

$=(a, \rho^*(b)) \quad by$ 

$=(\rho(a), b) \quad by$ 

$=\left((\rho(a), \cdot), (b, \cdot) \right) \quad by$ 

$=\left((a, \rho^*(\cdot)), (b, \cdot) \right) \quad by$ 

$=\left((\rho^*)^\dagger (a, \cdot), (b, \cdot) \right) \quad by$ 

So… its kind of like an Eckmann-Hilton argument. Anyhow, these ideas don’t work nicely in the case of 2-Hilbert spaces. That is, in the context when $A, B : H_1 \rightarrow H_2$ are maps of 2-Hilbert spaces, and $\rho : A \Rightarrow B$ is a 2-morphism, and we’re trying to compare the 2-morphisms

(2)$(\rho^\dagger)^*, (\rho^*)^\dagger : A^* \rightarrow B^*,$

where $A^*$ and $B^*$ are left adjoints of $A$ and $B$ respectively.

There are two problems : one on the “$*$” side and one on the “$\dagger$” side:

1. Whereas the dual $A^*$ of a vector space has a canonical definition $A^* = Hom (A, \mathbb{C})$, the dual (left adjoint) of a map $A : H_1 \rightarrow H_2$ of 2-Hilbert spaces has no basis-free definition… that I know of, at least.

2. There is no nice “Riesz representation theorem” way to write down the 2-morphism $\rho^\dagger : B \Rightarrow A$. Indeed, there are even two candidates for the definition, one using the left adjoint and the other the right adjoint, as I drew in string diagrams above. [Of course, there is a categorified Riesz representation theorem. Unlike the ordinary one though, it doesn’t give you an explicit formula .]

Mmmm…

Posted by: Bruce Bartlett on March 18, 2007 2:52 PM | Permalink | Reply to this

### Re: Star-Structures and Daggers

Bruce wrote:

Lemma : If $\rho : A \rightarrow B$, then $(\rho^\dagger)^* = (\rho^*)^\dagger$.

Nice proof!

But then you claim…

Anyhow, these ideas don’t work nicely in the case of 2-Hilbert spaces.

Maybe, maybe not. You’ll definitely need to mess around a lot, since everything is categorified one notch, and you’re trying to prove the formula for 2-morphisms instead of morphisms. But, something beautiful should work… and your objections don’t seem convincing to me. For example:

Whereas the dual $A^*$ of a vector space has a canonical definition $A^* = Hom(A,\mathbb{C})$, the dual (left adjoint) of a map $A: H_1 \to H_2$ of 2-Hilbert spaces has no basis-free definition… that I know of, at least.

Eh? The left adjoint of $A: H_1 \to H_2$ is the functor $A^*: H_2 \to H_1$ uniquely characterized (up to natural isomorphism) by the existence of a natural isomorphism

$hom(A^* w, v) \cong hom(A w, v)$

for all $v \in H_1$, $w \in H_2$.

By the way, this is awfully reminiscent of one of the formulas you were using in the proof of your Lemma.

Alas, it’s dinner-time, so I can’t think about this more now. I haven’t thought about this stuff for a long time, but one thing I know: it’s so simple and nice that nothing bad can possibly happen! There are some subjects in math where things get tricky and nasty, but linear algebra isn’t one.

Posted by: John Baez on March 19, 2007 4:54 AM | Permalink | Reply to this

### Re: Star-Structures and Daggers John wrote:

Eh? The left adjoint of $A : H_1 \rightarrow H_2 is the functor A^*:H_2 \rightarrow H_1$ uniquely characterized (up to natural isomorphism) by the existence of a natural isomorphism $hom(A^*w, v) \cong hom(Aw,v)$.

You’re right. But in some sense this is slightly unsatisfactory, in this context. When I said that the left adjoint $A^*$ to $A$ had no basis-free definition, I really meant that there was no explicit formula for it… defining a thing up to natural isomorphism is not really a formula , in the strict sense of the word, at least . I mean, its not even a definition, in the strictest sense… its really a property.

Its a bit different in the case of the usual adjoint $A^* : H_2 \rightarrow H_1$ of a linear map between Hilbert spaces $A : H_1 \rightarrow H_2$. In this case, $A^*$ is given uniquely… though I must admit, I know of no explicit formula for it either . Can anyone help?

For my benefit, as much as anyone else’s, I’ll recall how that works. You first prove the Riesz representation theorem.

Riesz representation theorem : If $f : H \rightarrow \mathbb{C}$ is a continous linear functional on a Hilbert space $H$, then $f = (v_f, \cdot)$ for a unique $v_f \in H$.

Proof : Assume $f \neq 0$. Choose a nonzero $z_f \in ker(f)^\perp$, which is a one-dimensional subspace of $H$. The choice

(1)$v_f = \frac{|f(z)|^2}{f(z) \|z\|^2} z_f.$

does the job. Note that its invariant under rescaling of $z_f$, so its unique.

Corollary : If $A : H_1 \rightarrow H_2$ is a continuous linear map, then there exists a unique map $A^* : H_2 \rightarrow H_1$ such that $(A^*w, v)_1 = (w, Av)_2$.

Proof : Given $w \in H_2$, define $f_w : H_1 \rightarrow \mathbb{C}$ by

(2)$f_w (v) = (w, Av)_2.$

Then by the Riesz rep theorem,

(3)$f_w (\cdot) = (v_{f_w}, \cdot)_1$

and we define

(4)$A^* (w) = v_{f_w}.$

Thats the usual way it goes, but its not really a formula , since it relies on us choosing a $v \in H_1$ such that $v$ is orthogonal to every $v' \in H$ such that $(w, Av')_2 = 0$, and then normalizing the result appropriately. The final result is unique, but I don’t know of an explicit formula. Then again, I guess thats the best one can hope for, when so little information is given.

After all this rambling, let me summarize my main points :

1. There isn’t an explicit formula for the adjoint of a functor between 2-Hilbert spaces. There can’t be, after all its only given up to unique isomorphism. But this messes one about when you try to prove that $(A^\dagger)^* = (A^*)^\dagger$.

2. This motivates one to differ slightly from the philosophy of a “monoidal 2-category with duals” as presented by John Baez and Laurel Langford in their 2-tangles paper. There they say, in essence, that duality is completely a structure. For every 2-morphism $\theta : F \Rightarrow G$ there is given a dual 2-morphism $\theta^* : G \Rightarrow F$, for every morphism $F : H_1 \rightarrow H_2$ there is given a dual morphism $F^* : H_2 \rightarrow H_1$ and every object $H$ has a given dual $H^*$.

I would prefer to think of duality only at the top-level as structure, and the rest as properties. In other words, every 2-morphism $\theta : F \Rightarrow G$ has a given dual $\theta^* : G \rightarrow F$; for every morphism $F : H_1 \rightarrow H_2$ there exists a $F^* : H_2 \rightarrow H_1$ such that $(F^*w, v)_1 \cong (w, Fv)_2$, etc.; and for every object $H$ there exists a dual $H^*$ such that, etc., etc.

What do you think?

Posted by: Bruce Bartlett on March 25, 2007 2:01 PM | Permalink | Reply to this
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