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I have a rather basic question about $C^*$-Frobenius algebras (also called Q-systems). Any pointers or references will be most helpful!

We are given a finite-dimensional complex Hilbert space $\mathbb{V}$ with a multiplication $m: \mathbb{V} \otimes \mathbb{V} \rightarrow \mathbb{V}$ (a linear map) such that

  1. $m$ is associative
  2. $m^\dagger m = \mathbb{I}$ (the Identity on $\mathbb{V}$), that is, $m$ is an isometry, and
  3. $m$ and $m^\dagger$ fulfill the Frobenius relation, namely, $(m^\dagger \otimes \mathbb{I}) (\mathbb{I} \otimes m) = m m^\dagger$,

[Here $m^\dagger$ is the Hermitian conjugate (adjoint) of $m$. If $m$ is viewed as a matrix from $\mathbb{V} \otimes \mathbb{V}$ to $\mathbb{V}$ then $m^\dagger$ is obtained by first transposing this matrix and then applying entry-wise complex conjugation.]

These relations are depicted by the string diagrams shown below:

String diagrams for the relations (1)-(3)

My question is: Given this data, does the multiplication $m$ necessarily have a unit? If yes, can it be expressed in terms of $m$ and $m^\dagger$?

Improve this question
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  • $\begingroup$ What precisely is the Frobenius relation? In fact, it would be nice if you carefully specified what the other entities in your question are. For example, I assume your tensor product is the Hilbert space tensor product, and that your adjoint should mean the adjoint of a bounder or at least closed linear operator between the two Hilbert spaces. Is your multiplication a linear map? All of these things should be specified in the body of the question, unless you can point to a clear reference to your definition. $\endgroup$
    – Jon Bannon
    Jul 24, 2020 at 16:14
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    $\begingroup$ Hi Jon, I've added some more information. I'm just thinking of finite-dimensional Hilbert spaces with the usual tensor product and the multiplication as a linear map (as a matrix from $\mathbb{V} \otimes \mathbb{V}$ to $\mathbb{V}$). $\endgroup$
    – quantumOrange
    Jul 24, 2020 at 16:35
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    $\begingroup$ In 2. don't you mean $m m^{\dagger}= Id$? $m$ cannot be an isometry unless the dimension of $V$ is equal to $1$. The Frobenius identity, as you write, cannot possibly hold because the left-hand side and the right-hand side act on different spaces. $\endgroup$
    – Mateusz Wasilewski
    Jul 24, 2020 at 17:13
  • $\begingroup$ Hi Mateusz, to clarify my convention I've now added string diagrams of these relations. (These diagrams are read from top to bottom.) $\endgroup$
    – quantumOrange
    Jul 27, 2020 at 8:03

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