Fusion rings

I'll more or less stick to the presentation given in this question: [1]

We define a fusion ring as follows: consider a free $\mathbb{Z}$-module $\mathbb{Z}\mathcal{B}$ with finite basis $\mathcal{B}=\{b_{1},b_{2},\cdots,b_{n}\}$. Equip this module with a binary product such that we get a $\mathbb{Z}$-algebra $\mathcal{F}=(\mathbb{Z}\mathcal{B},\cdot)$ where $b_{1}$ acts as a multiplicative identity in the ring $\mathcal{F}$ and \begin{equation*}b_{i}\cdot b_{j}=\sum_{k}N_{ij}^{k} \ b_{k} , \quad N_{ij}^{k}\in\mathbb{N}_{0}\end{equation*} $\mathcal{F}$ is called a fusion ring/algebra. In particular, note that $N_{i1}^{j}=N_{1i}^{j}=\delta_{ij}$ (multiplicative identity).

We add the following bit of structure ('invertibility'): for every $b_{i}\in\mathcal{B}$, there exists some unique $b_{j}\in\mathcal{B}$ such that $b_{1}$ occurs in the decomposition of $b_{i}\cdot b_{j}$ and $b_{j}\cdot b_{i}$. Denote this 'inverse' by $b_{i^{*}}$. That is, \begin{equation*}\forall i \ \ \exists ! j \ : N_{ij}^{1}=N_{ji}^{1}>0\end{equation*} where \begin{equation*}N_{i^{*}k}^{1}=N_{ki^{*}}^{1}=\delta_{ik}\end{equation*}


Referring back to [1], it is asserted (in the comments) that

\begin{equation*}N_{ij}^{k}=N_{j^{*}i^{*}}^{k^{*}}\end{equation*} for all $i,j,k$.

Question: For a fusion algebra with the above structure (neutrality, invertibility and associativity of product), what's the proof (or a reference for one) for this identity?

I've tried a few things but can't quite seem to get it. The result does appear to follow if we can guarantee that $*:i\mapsto i^{*}$ defines an anti-isomorphism of a fusion algebra.

EDIT: After a bit of searching around, it appears that a few places seem to include "$*$" inducing an anti-isomorphism of $\mathcal{F}$ as part of the definition. I suppose that can be motivated by $b_{1}$ being guaranteed to be contained in \begin{equation*}(b_{i}b_{j})(b_{j^{*}}b_{{i}^{*}}) \quad \text{and} \quad (b_{i}b_{j})(b_{j^{*}}b_{{i}^{*}})\end{equation*}

Still, I wonder if this is absolutely necessary...

  • $\begingroup$ $b_{i}\cdot b_{j}=b_{1}=b_{j}\cdot b_{i}$ is much stronger than what you expected because it implies that $d(b_i)d(b_j)=1$, so that $d(b_i) = d(b_j) = 1$, which means that $\mathcal{B}$ forms a finite groups. Now what you wrote just after is the correct statement. $\endgroup$ – Sebastien Palcoux Feb 6 at 6:08
  • $\begingroup$ All the identities you are talking about reduce to a single one called the Frobenius reciprocity: $$N_{i,j}^k = N_{i^*,k}^j = N_{k,j^*}^i.$$ In particular what you call identity II can be proved directly from the Frobenius reciprocity as follows: $$N_{i,j}^k = N_{i^*,k}^j = N_{j,k^*}^{i^*} = N_{j^*,i^*}^{k^*}.$$ So you should focus on how to prove the Frobenius reciprocity. $\endgroup$ – Sebastien Palcoux Feb 6 at 6:14
  • 1
    $\begingroup$ We can prove that (for a fusion ring) the Frobenius reciprocity is equivalent to $∗$ being an antihomomorphism of algebra, but (you are right) I don't know if it can be proven independently. So, pending a relevant answer to this problem, I have decided to modify the post of me you cited by including the Frobenius reciprocity into the definition of a fusion ring. $\endgroup$ – Sebastien Palcoux Feb 6 at 6:51
  • $\begingroup$ You should completely rewrite your post as follows: first fix the mistake with "invertibility". Next ask whether "Associativity", "Neutral" and "Inverse/Adjoint" (as written in my post) are sufficient to prove "Frobenius reciprocity" (or if you prefer, that $*$ is an antihomomorphism of algebra). $\endgroup$ – Sebastien Palcoux Feb 6 at 6:56
  • 1
    $\begingroup$ What you wrote: $$\forall i \ \ \exists ! i^{*} \ : N_{i,i^{*}}^{1}=1=N_{i^{*},i}^{1}$$ is also not sufficient. You should write: $$\forall i \ \ \exists ! j \ : N_{i,j}^{1} > 0,$$ such one is denoted $i^∗$. Next, $N_{i^*,k}^{1} = N_{k,i^*}^{1} = \delta_{i,k}$ (should these last equalities be also in the assumption, or be a renormalization, or be a deduction?). $\endgroup$ – Sebastien Palcoux Feb 6 at 7:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.