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$\newcommand{\id}{\mathrm{id}}$My reference for this post is Musto, Reutter and Verdon's A compositional approach to quantum functions, arXiv:1711.07945. Questions are in bold below. Allow me to begin with a bit of background. Taking inspiration from Gelfand duality for commutative $C^*$-algebras, we take the notion of a quantum set to be a Frobenius algebra, which is defined to be a finite dimensional Hilbert space $H$, which is a unital algebra, with multiplication $\mu : H\otimes H\to H$ and unit $\eta : \mathbb{C}\to H$, that has a co-multiplication $\delta : H\to H\otimes H$ and co-unit $\epsilon : H\to \mathbb{C}$, and satisfies the Frobenius condition: $$(\mu\otimes \id)(\id\otimes\delta) = \delta\circ\mu = (\id\otimes \mu)(\delta\otimes \id).$$ A SSFA is a special symmetric Frobenius algebra (which is just a special class of Frobenius algebras (see the above arXiv reference)). Given a linear function $f : A\to B$ between SSFA's $A$ and $B$, the dagger $f^\dagger : B^*\to A^*$ is just the Hilbert space adjoint of $f$. Then $\delta = m^\dagger$ and $\epsilon = \eta^\dagger$.

There is a diagrammatic approach to the formulation a Frobenius algebra, which involves replacing the algebraic axioms of multiplication, unitality, co-multiplication, etc, with nice string diagrams. enter image description here What I am concerned about lies in the image below. enter image description here Question 1: How do we interpret the formulation of the $C^*$-involution in Theorem 2.6 algebraically (diagram (15))?

Question 2: How do we interpret the diagram for the "dagger formula" of a $*$-homomorphism / $*$-co-homomorphism in Definition 2.7 algebraically (third diagram on the right in (16) / (17))?

Question 2 is what I'm really after, but I figure the main gap is my understanding of question 1. The multiplicative / co-multiplicative and unital / co-unital diagrammatic properties in Definition 2.7 are clear to me. The diagram in question for Definition 2.7 should correspond to the involution property of a $*$-morphism: $f(x^*) = f(x)^*$. I am not seeing it when I try to convert the diagram into algebra. Here is may failed attempt.

The upside popsicle diagram in Theorem 2.6 looks to me like it should be the "point map" $a : \mathbb{C}\to A$, $\lambda\mapsto \lambda a$. Then $a^\dagger : A^*\to \mathbb{C}$ is the adjoint of the map $a$, and replacing the strings with their definitions in terms of the co-multiplication and unit, we get that the involution is $$a\mapsto (\id\otimes a^\dagger)\delta(1) = 1\otimes a^\dagger(1) = a^\dagger(1)\in \mathbb{C}.$$ This is not an antiautomorphism of $A$ ,so, obviously, I am quite confused about these diagrams.

I am knowledgeable on Woronowicz's formulation of a compact quantum group and its operator algebraic aspects. I have some (albeit limited) knowledge on Frobenius algebras defined in terms of Frobenius forms and in terms of algebras and co-algebras. More generally, I am knowledgeable on the basics of operator algebras. My main my goal with this paper is to understand the notion of a quantum function and a quantum element of a quantum set. In particular, I find Proposition 4.12 intriguing. It seems like it's possibly a non-commutative analogue of the fact classical permutations are just bijections on a classical finite set. For this, I am required to translate the concepts into a language I understand.

Any help is much appreciated!

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    $\begingroup$ For Q1: a Frobenius algebra is not a bialgebra, so I'm not sure if $\delta(1)$ equals $1\otimes1$ (which the mixes algebra and coalgebra structures). In Sweedler notation $\delta(1)=\sum 1_{(1)} \otimes 1_{(2)}$ so we seem to get $a\mapsto \sum a^\dag(1_{(2)}) \, 1_{(1)}$ which still $\in A$. [Also, I would think the last diagrams in (12) and (17) ought to be reflected] $\endgroup$
    – Jules Lamers
    Commented Dec 14, 2021 at 23:13
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    $\begingroup$ .. And I suppose that in that way Thm 2.6 gives an inner product, allowing us to switch between elements and their duals. I think that on the right in (16) the author omitted the little 'popsicle diagram' for the unit dangling at the bottom in (15). So for Q2 I think that the relation says: $f^\dag(x) = (f(x^\dag))^\dag$. $\endgroup$
    – Jules Lamers
    Commented Dec 14, 2021 at 23:28
  • $\begingroup$ Oh, wow, yes, you're absolutely right. I'm mixing up the co-algebra and algebra structures. Thanks for your help! $\endgroup$
    – Ben A-S
    Commented Dec 15, 2021 at 2:34
  • $\begingroup$ Is it true that a Frobenius algebra is trivial if and only if $\delta(1) = 1\otimes 1$? $\endgroup$
    – Ben A-S
    Commented Dec 15, 2021 at 2:58
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    $\begingroup$ I also find it hard to work "just with diagrams". Vicary showed in arxiv.org/abs/0805.0432 that SSFA's are just finite-dimensional $C^*$-algebras equipped with a trace satisfying a certain normalisation condition (which I always forget...) The trace turns the $C^*$-algebra into a Hilbert space, and $\delta$ is the the Hilbert space adjoint of the multiplication map. This perspective makes it clear that $\delta(1)=1\otimes 1$ only for trivial $A$. $\endgroup$
    – Matthew Daws
    Commented Dec 15, 2021 at 8:36

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