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Stream: theory: category theory

Topic: Dagger compact category from bilinear form

Jean-Baptiste Vienney (Jun 08 2023 at 02:27):

Let $(\mathcal{C},\otimes,I,\_^*)$ be a compact closed category ie. a symmetric monoidal category with a functor $\_^*:\mathcal{C}^{op} \rightarrow \mathcal{C}$ and evaluation/co-evaluation natural transformations. Suppose also given a natural transformations $s_{X}:X \otimes X \rightarrow I$ (the idea is that it could be a scalar product). Then by using the compact closure, I get a natural transformation $i_{X}:X^* \rightarrow X$ and also $j_{X}:X \rightarrow X^*$. Now starting with $f:X \rightarrow Y$, I get $f^{*}:Y^* \rightarrow X^*$ and then $j_{Y};f^*;i_{X}:Y \rightarrow X$. My idea is that by imposing appropriate conditions on $s$, I should obtain by defining $f^{\dag} = j_{Y};f^*;i_{X}$, that $(\mathcal{C},\otimes,I,\_^*,\_^\dagger)$ is a dagger compact category.

I have two questions:
1) Does anyone have already thought to that?
2) If that's not the case, should I work for trying making it work? I mean do you find this idea interesting and valuable?

John Baez (Jun 08 2023 at 02:51):

Are you demanding that $s_X$ be natural in $X$? Then the category of finite-dimensional real inner product spaces and arbitrary linear maps will not be a category of this form, because arbitrary linear maps don't preserve the inner product. At first I thought the category of finite-dimensional real inner product spaces and inner-product preserving linear maps would be of this form, but then the problem is that this category is not compact closed, because the evaluation map $\epsilon_X : X^\ast \otimes X \to I$ does not preserve the inner product (because it's usually not injective).

John Baez (Jun 08 2023 at 03:05):

But I think there are ways to fix this idea.

It might be good to talk to some people in the Oxford crowd, who study dagger compact categories, to see if this idea has been studied. For example you could ask @Chris Heunen and @Jamie Vicary, who wrote a nice book Categories for Quantum Theory, with material on "dagger dual objects".

Jean-Baptiste Vienney (Jun 08 2023 at 03:07):

John Baez said:

Are you demanding that $s_X$ be natural in $X$? Then the category of finite-dimensional real inner product spaces and arbitrary linear maps will not be a category of this form, because arbitrary linear maps don't preserve the inner product. At first I thought the category of finite-dimensional real inner product spaces and inner-product preserving linear maps would be of this form, but then the problem is that this category is not compact closed, because the evaluation map $\epsilon_X : X^\ast \otimes X \to I$ does not preserve the inner product (because it's usually not injective).

In fact $s_{X}$ should not be natural you're right, I remarked a few minutes ago that it doesn't make sense.

Jean-Baptiste Vienney (Jun 08 2023 at 03:10):

John Baez said:

But I think there are ways to fix this idea.

It might be good to talk to some people in the Oxford crowd, who study dagger compact categories, to see if this idea has been studied. For example you could ask Chris Heunen and Jamie Vicary, who wrote a nice book Categories for Quantum Theory, full of ideas on compact dagger categories.

Ok, let's see if they see this first so.

John Baez (Jun 08 2023 at 03:12):

It's good to email them, since I don't think they visit this site very often, especially not Jamie.

John Baez (Jun 08 2023 at 03:13):

It's also good to get ahold of that book.

Jean-Baptiste Vienney (Jun 08 2023 at 04:27):

I did these two things.

Mike Shulman (Jun 08 2023 at 04:31):

Are you hoping that the "appropriate conditions" on $s$ would be something weaker than assuming that $i$ and $j$ are inverse isomorphisms? It's not clear to me how else you could ensure $f^{\dagger\dagger}=f$.

If $i$ and $j$ are inverse isomorphisms, then you essentially have a symmetric monoidal category in which every object is self-dual, and it makes sense to me that that would automatically be dagger-compact.

Jean-Baptiste Vienney (Jun 08 2023 at 04:42):

I didn't think that it would be necessary to assume that $i$ and $j$ are inverse isomorphisms but maybe you are right and that's the only solution. By the way do you know dagger compact categories such that $A$ is not isomorphic to $A^*$? I only know $FHilb$ and $Rel$ and they are such that every object is isomorphic to its dual.

Mike Shulman (Jun 08 2023 at 04:53):

No, I don't know any dagger compact categories where not all objects are self-dual.

I observe that the fact that self-duality in a symmetric monoidal category makes it dagger-compact is mentioned on the nLab page [[dagger-compact category]].

Jean-Baptiste Vienney (Jun 08 2023 at 05:10):

Ok, but maybe it is not necessary to assume that $i$ and $j$ are inverse isomorphisms. We should understand what $f^{\dag\dag} = f$ is equivalent to when you define $f^{\dagger}$ this like this.

Jean-Baptiste Vienney (Jun 08 2023 at 05:13):

I mean, that's the only way to understand this question.

Jean-Baptiste Vienney (Jun 08 2023 at 05:22):

But $Id_X^{\dag} = Id_X \Rightarrow j_{X};i_{X}=Id_{X}$

Jean-Baptiste Vienney (Jun 08 2023 at 11:48):

That was a good idea to email them. I started to discuss with Chris. By the way, I've already met Chris last year so I was less scared to reach out to him than to Jamie. One new thing I understood by replying to the reply of Chris is that to define $i_{X}$ and $j_{X}$ you only need coevalutation, and then you can obtain a map of type eval from $j$ and $s$.

John Baez (Jun 08 2023 at 23:37):

Hmm

Mike Shulman said:

No, I don't know any dagger compact categories where not all objects are self-dual.

Yikes! I should know some, but do I? Already with finite-dimensional complex Hilbert spaces we notice something funny: though every object is isomorphic to its dual, the isomorphism is not "natural" (in some intuitive sense), because the map

$H \to H^\ast$

sending $v \in H$ to $\langle v , - \rangle \in H^\ast$ is not linear: it's conjugate-linear.

(Here I'm using physics conventions, where $\langle v, w \rangle$ is conjugate-linear in $v$ and linear in $w$.)

John Baez (Jun 08 2023 at 23:38):

This makes me feel some fancier category, built using finite-dimensional complex Hilbert spaces, should have the property that it's a compact dagger category but not every object is isomorphic to its dual!

John Baez (Jun 08 2023 at 23:39):

Oh, right, it's obvious: the category of finite-dimensional continous unitary representations of the group $\mathrm{SU}(3)$.

John Baez (Jun 08 2023 at 23:40):

This group has two 3-dimensional irreducible representations, the "tautologous" one and its dual, which are not isomorphic.

John Baez (Jun 08 2023 at 23:40):

I was joking when I said this is "obvious", but in physics this is why quarks are different than antiquarks!

John Baez (Jun 08 2023 at 23:41):

A less showoffy example would be the category of finite-dimensional continuous unitary representations of the circle group, $\mathrm{U}(1)$.

John Baez (Jun 08 2023 at 23:42):

But basically I had to turn off the category theory part of my brain and turn on the physics part of my brain to solve this puzzle.

Mike Shulman (Jun 09 2023 at 00:39):

Nice! It would be great to have this example at [[dagger compact category]]...

Jean-Baptiste Vienney (Jun 09 2023 at 00:40):

Cool! so now I changed a bit my mind. What I would want to say is that all you need to define a dagger compact category is for every object $X$, an object $X^*$ and two maps $s_{X}:X \otimes X \rightarrow I$ and $coev:I \rightarrow X \otimes X^*$ which verify "some conditions". By drawing a bit of diagrams, you can find that it is enough to get some maps of type $X \otimes X^* \rightarrow I$ and to transform a map $f:X \rightarrow Y$ into something $Y \rightarrow X$. It makes obviously less structure in the definition but I foresee that in counterpart the equations to be verified will be more complicated.

Jean-Baptiste Vienney (Jun 09 2023 at 00:42):

Maybe it could define something slightly less general that a dagger compact category, like a specific class of them, but it could be closer to usual physics as it would use a map of type "scalar product".

John Baez (Jun 09 2023 at 01:30):

$s_X : X \otimes X \to X$ is a typo, right? Did you mean $s_X : X^\ast \otimes X \to I$?

Jean-Baptiste Vienney (Jun 09 2023 at 01:31):

No I want $X \otimes X \rightarrow I$!

John Baez (Jun 09 2023 at 01:33):

Oh. Believe it or not, that's what I was trying to type!

John Baez (Jun 09 2023 at 01:34):

But maybe this is a more efficient way to say what you're trying to say: we've got a compact closed category and for each object $X$ we have a chosen isomorphism $f: X \to X^\ast$. That will give us a dagger compact category.

John Baez (Jun 09 2023 at 01:34):

Or do you really want your formalism to handle more general situations?

Jean-Baptiste Vienney (Jun 09 2023 at 01:35):

I would want to handle more general more situations.

Jean-Baptiste Vienney (Jun 09 2023 at 01:35):

But the axioms of a dagger compact category written in this situation by using these two type of maps looks horrible.

Jean-Baptiste Vienney (Jun 09 2023 at 01:36):

You need to use dozen of caps and cups to write them.

John Baez (Jun 09 2023 at 01:36):

Can you give an example of one of these more general situation? I'm having trouble understanding what examples you're trying to handle.

Jean-Baptiste Vienney (Jun 09 2023 at 01:36):

Situations like what you gave for example.

Jean-Baptiste Vienney (Jun 09 2023 at 01:37):

I don't want to handle to handle more situations that a dagger compact category. What I wanted was just to understand how they arise from a map of the type of a scalar product.

John Baez (Jun 09 2023 at 01:37):

But my examples had no morphisms $s_X: X \otimes X \to I$ except the zero morphism.

Jean-Baptiste Vienney (Jun 09 2023 at 01:38):

Ok, so that's maybe just uninteresting. You don't have a scalar product?

Jean-Baptiste Vienney (Jun 09 2023 at 01:39):

Or just some bilinear form.

Jean-Baptiste Vienney (Jun 09 2023 at 01:44):

Ok, I believe you.

Jean-Baptiste Vienney (Jun 09 2023 at 01:45):

These categories look very small, that's funny.

Jean-Baptiste Vienney (Jun 09 2023 at 01:49):

By the way, that's always great when you understand that you didn't understand something very well and it opens your eyes to new things to learn.

John Baez (Jun 09 2023 at 02:01):

It's fun to study dagger categories along with group representation theory since they both grew out of quantum mechanics and the category of finite-dimensional continuous unitary representations of a compact Lie group is a dagger compact category. A good example of a compact Lie group is the circle group

$\mathrm{U}(1) = \{e^{i\theta} : \theta \in \mathbb{R} \} \subset \mathbb{C}^\ast$

This has one irreducible representation for each $n \in \mathbb{Z}$; they're all representations on $\mathbb{C}$ and on the nth one of these representations $e^{i\theta}$ acts as multiplication by $e^{i n \theta}$. Let's call the nth representation $\rho_n$. Then we have

$\rho_n \otimes \rho_m \cong \rho_{n + m}$

and $\rho_0$ is the trivial representation, the unit for the tensor product. So if you think a bit you can see

$\rho_n^\ast \cong \rho_{-n}$

Also by Schur's lemma there are no morphisms except the zero linear map between $\rho_n$ and $\rho_m$ when $n \ne m$.

This is a nice dagger compact category where the objects aren't isomorphic to their duals and the only morphism $\rho_n \otimes \rho_n \to I$ is the zero morphism, except when $n = 0$.

Jean-Baptiste Vienney (Jun 09 2023 at 02:13):

Nice, thank you! I'm excited to learn about group representation theory now.

Jean-Baptiste Vienney (Jun 09 2023 at 02:14):

I was already excited to learn about representations of the symmetric group but now, I'm also excited by representations of the other groups.

John Baez (Jun 09 2023 at 02:19):

Yes, by the way you don't need to go to Lie groups to find groups that have representations not isomorphic to their duals!

John Baez (Jun 09 2023 at 02:20):

For the symmetric groups, all representations are isomorphic to their duals.

John Baez (Jun 09 2023 at 02:21):

But if you check out $\mathbb{Z}/3$, you'll see it has 3 different irreducible representations: the trivial rep and two other 1d reps, which are duals of each other.

Jean-Baptiste Vienney (Jun 09 2023 at 02:22):

John Baez said:

For the symmetric groups, all representations are isomorphic to their duals.

Maybe only in characteristic 0!

John Baez (Jun 09 2023 at 02:22):

Yes, I certainly meant in characteristic zero; I was actually thinking about complex reps in all my remarks just now!

John Baez (Jun 09 2023 at 02:23):

For example the stuff I said about representations of $\mathbb{Z}/3$ would be false over the rationals.

John Baez (Jun 09 2023 at 02:23):

Over the rationals $\mathbb{Z}/3$ has just two irreducible representations!

John Baez (Jun 09 2023 at 02:25):

And over the rationals every rep of a finite group is self-dual.

Jean-Baptiste Vienney (Jun 09 2023 at 02:26):

Ok, I will try to prove this by myself to exercise.

John Baez (Jun 09 2023 at 02:26):

Yes, $\mathbb{Z}/3$ is a very good group to think about.

John Baez (Jun 09 2023 at 02:28):

Representations of $\mathbb{Z}/2$ are a bit boring compared to those of $\mathbb{Z}/3$. When you work a bit you'll see this is because every square root of 1 is rational, but not every cube root of 1 is rational.

John Baez (Jun 09 2023 at 02:29):

Working out the rational, real and complex irreps of $\mathbb{Z}/3$, you'll secretly start learning about "splitting fields".

Chris Heunen (Jun 09 2023 at 08:25):

Similar ideas are discussed in [Knus, "Quadratic and Hermitian forms over rings", chapter 2] and [Scharlau, "Quadratic and Hermitian forms", chapter 7]. They mostly consider abelian categories, and study extra properties on the form such as anisotropy to make them more well-behaved. But they don't use compactness but only a contravariant endofunctor * on the category with a natural isomorphism between ** and the identity. For the construction you're proposing to be a (functorial and involutive) dagger, I think you do need the evaluation, and moreover not just a choice of form s_X on each object X, but also coherence between s_X and s_{X* } - roughly, i needs to be the inverse of the transpose of j. So you may also be interested in [Selinger, "Autonomous categories in which A=A*"].

[By the way, I'm indeed not often on here, but please don't be afraid to contact anyone! @Matthew Di Meglio and @Martti Karvonen may also be interested.]