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Stream: learning: questions

Topic: dagger categories

Faré (Apr 13 2020 at 21:06):

Are there cases where $C^\dagger$ and $C^{op}$ are both defined but mean different things? If not, can the two notations be unified? If yes, what are interesting examples?

Evan Patterson (Apr 13 2020 at 21:56):

The dagger is an operation on morphisms in a special kind of category (a dagger category), whereas the opposite is an operation on categories. So they are not directly comparable.

There are dagger categories that are not equivalent to their opposites, which is possibly what you are getting at. This is true of the prototypical example of the category of Hilbert spaces and linear maps, where the dagger is the adjoint.

Amar Hadzihasanovic (Apr 13 2020 at 22:02):

A dagger is a contravariant endofunctor that squares to the identity, so every dagger category is not only equivalent, but isomorphic to its opposite.

Maybe you are thinking of something else, eg the "dual" functor on Banach spaces?

Evan Patterson (Apr 13 2020 at 22:04):

They are isomorphic only in a contravariant sense. I guess I always understood isomorphism to be covariant unless explicitly stated.

Amar Hadzihasanovic (Apr 13 2020 at 22:06):

What is "a covariant isomorphism between a category and its opposite"?

Evan Patterson (Apr 13 2020 at 22:10):

It is a (covariant) invertible functor $C^{op} \to C$, which is (by definition) a contravariant functor $C \to C$. Maybe I am missing your point.

Evan Patterson (Apr 13 2020 at 22:11):

In my answer, I was trying to explain how the category Hilb is not like the category Rel.

Amar Hadzihasanovic (Apr 13 2020 at 22:14):

I am saying that a dagger is exactly that, and that in that respect Hilb is exactly like Rel.

Faré (Apr 13 2020 at 22:14):

The example of dagger in Hilbert spaces being adjoint is helpful. But I'm still confused as to how confused I may or may not be. Is $Cat$ a dagger category with ${\_}^{op}$ as its dagger? If not, why not?

Matt Feller (Apr 13 2020 at 22:15):

I think co/contravariant are distracting terms to get caught up in here. The only way you would want to talk about functors from a category to opposite is to think of them as covariant because a contravariant functor from a category to its opposite amounts to a covariant functor from a category to itself.

James Wood (Apr 13 2020 at 22:17):

† applies to morphisms, whereas ᵒᵖ applies to objects of Cat.

Amar Hadzihasanovic (Apr 13 2020 at 22:17):

@Faré A dagger is the identity on objects, inverts the direction of morphisms, and squares to the identity.

$-^{op}$ is not a dagger on Cat because it is not the identity on categories and it does not invert the direction of functors.

Matt Feller (Apr 13 2020 at 22:18):

Faré said:

The example of dagger in Hilbert spaces being adjoint is helpful. But I'm still confused as to how confused I may or may not be. Is $Cat$ a dagger category with ${\_}^{op}$ as its dagger? If not, why not?

When you apply op to a functor $C\to D$, it gives you a functor $C^{op}\to D^{op}$ instead of $D\to C$

Amar Hadzihasanovic (Apr 13 2020 at 22:18):

It does however invert the direction of natural transformations: it's "contravariant on 2-morphisms" in the 2-category Cat.

Faré (Apr 13 2020 at 22:19):

Also, where does it matter to make it $C^{op} \longrightarrow C$ versus $C \longrightarrow C^{op}$ ? When I see say $Id_A^{\dagger}$ my brain has trouble parsing the $A$ as being an opposite being righted rather than a normal thing being opposed.

I guess that as a computer scientist, I'm utterly baffled at the seemingly very complex type inference that mathematicians (and even worse, physicists) take as granted and free, yet that I trouble running in my head.

Faré (Apr 13 2020 at 22:22):

Thanks, makes much more sense now.

John Baez (Apr 13 2020 at 22:25):

It's a functor $F: C \to C^{\mathrm{op}}$ that has an inverse.

John Baez (Apr 13 2020 at 22:25):

An isomorphism between categories is a functor that has an inverse.

James Wood (Apr 13 2020 at 22:27):

A functor F : Cᵒᵖ → D contains a function ∀BA. Cᵒᵖ(B, A) → D(F B, F A), whereas a functor F : C → Dᵒᵖ contains a function ∀AB. C(A, B) → Dᵒᵖ(F A, F B). When you expand the definition of _ᵒᵖ, these are the same. The laws work out to be equivalent too.

sarahzrf (Apr 13 2020 at 23:00):

i dont like defining dagger categories as a functor

sarahzrf (Apr 13 2020 at 23:00):

it makes them sound evil when they arent

sarahzrf (Apr 13 2020 at 23:01):

or at least i don't think they are? i guess it depends on exactly what you mean

sarahzrf (Apr 13 2020 at 23:06):

but you can rephrase the definition as

for each object A and object B, a function †_A,B : Hom(A, B) → Hom(B, A), such that:

• for all A, we have †_A,A(id_A) = id_A
• for all A -f-> B -g-> C, we have †_A,C(g ∘ f) = †_A,B(f) ∘ †_B,C(g)
• †_A,B and †_B,A are inverse

sarahzrf (Apr 13 2020 at 23:08):

there is no reference here to equality of objects!

sarahzrf (Apr 13 2020 at 23:12):

...huh. https://mathoverflow.net/a/220108

Faré (Apr 13 2020 at 23:39):

Thank you all for bearing with me. This mathoverflow question was great, especially the "undirected category" isomorphism.

Max New (Mar 25 2022 at 21:22):

Are dagger categories known to be some kind of enriched, internal, or generalized multi- or poly-category? The description on that mathoverflow question definitely has that flavor

Mike Shulman (Mar 25 2022 at 21:30):

The closest I've ever come to something like this is identifying a bicategory of "dagger-matrices" in which dagger-categories should be the monads. A dagger-matrix is a matrix (of sets) equipped with an isomorphism to its transpose. Unfortunately any dagger-matrix must be square, so the bicategory of dagger-matrices is just a disjoint union of monoidal categories. It gets a little more interesting when you beef it up to a double category, but the lack of non-endo profunctors makes it hard to do any of the formal category theory that you can usually do with monads in a bicategory or double category.

Max New (Mar 25 2022 at 21:44):

The double category should be enough to get the right definition of dagger functor and dagger transformation though? You just don't get very many dagger profunctors I guess

Mike Shulman (Mar 25 2022 at 22:39):

dagger functor, certainly. Maybe dagger transformation -- I'm not sure I know what a dagger transformation is.

Cole Comfort (Mar 26 2022 at 18:52):

On page 5 after definition 2.6 (https://arxiv.org/pdf/1803.06651.pdf) the authors remark,

There is no need to go further and define ‘dagger natural transformations, if $\sigma:F\Rightarrow G$ is a natural transformation between dagger functors, then taking daggers componentwise defines a natural transformation $\sigma^\dag :G\Rightarrow F$

Mike Shulman (Mar 26 2022 at 18:53):

Hmm. That makes me think that maybe there isn't even a double category.

Mike Shulman (Mar 26 2022 at 18:55):

A morphism of dagger-matrices should preserve the involutions, but if there are different vertical arrows on the right and left, it's not clear what that should mean.