AnswerBun.com

Does every special $C^*$-Frobenius algebra have a unit?

MathOverflow Asked by quantumOrange on January 1, 2022

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 $m^dagger$?

Add your own answers!

Related Questions

Euler function summation

0  Asked on December 13, 2020 by andrej-leko

 

Ask a Question

Get help from others!

© 2023 AnswerBun.com. All rights reserved. Sites we Love: PCI Database, MenuIva, UKBizDB, Menu Kuliner, Sharing RPP, SolveDir