# A fusion ring identity

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 $$\begin{equation*}N_{i^{*}k}^{1}=N_{ki^{*}}^{1}=\delta_{ik}\end{equation*}$$

Question

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...

• $b_{i}\cdot b_{j}=b_{1}=b_{j}\cdot b_{i}$ 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. – Sebastien Palcoux Feb 6 at 6:08
• 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. – Sebastien Palcoux Feb 6 at 6:14
• 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. – Sebastien Palcoux Feb 6 at 6:51
• 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). – Sebastien Palcoux Feb 6 at 6:56
• 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?). – Sebastien Palcoux Feb 6 at 7:50