To avoid any confusion, we rewrite the basic definitions for a fusion ring (already written in this post).

A

fusion ringis a finite dimensional complex space $\mathbb{C}\mathcal{B}$ together with a distinguished basis $\mathcal{B} = \{ h_1,...,h_r\}$ and fusion rules $ h_i \cdot h_j = \sum_k n_{ij}^kh_k $, with $n_{ij}^k \in \mathbb{N}_{\geq 0}$ satisfying:

-Neutral: $n_{1i}^j = n_{i1}^j = \delta_{ij}$

-Dual: $\forall i \ \exists!j $ (noted $i^*$) such that $n_{ij}^1>0$

-Associativity: $\sum_s n_{ij}^sn_{sk}^t = \sum_s n_{jk}^sn_{is}^t$

-Frobenius-Perron reciprocity: $n_{ij}^k = n_{i^*k}^j = n_{kj^*}^i$

Remark: $\mathbb{C}\mathcal{B}$ admits a structure of finite dimensional ${\rm C}^*$-algebra (take $h_i^* = h_{i^*}$).

Frobenius-Perron theorem: $\exists!$ $*$-homomorphism $d:\mathbb{C}\mathcal{B} \to \mathbb{C}$ with $d(\mathcal{B}) \subset (0,\infty)$.The

rank$r$ of the fusion ring $\mathbb{C}\mathcal{B}$ is the cardinal of $\mathcal{B}$.

It is is calledintegralif every $d(h_i)$ is an integer.

ItsFrobenius-Perron dimension(FPdim) is $\sum d(h_i)^2$.

It is ofFrobenius typeif every $d(h_i)$ divides FPdim$(\mathbb{C}\mathcal{B})$

It issimpleif $r>1$ and for any fusion subring $\mathbb{C}\mathcal{S} \subseteq \mathbb{C}\mathcal{B}$ with $\mathcal{S} \subseteq \mathcal{B}$, then $\mathcal{S} = \{ h_1 \}$ or $\mathcal{B}$.

Open problem: Every fusion ring is of Frobenius type.

Remark: The Grothendieck ring of a finite group $G$ is the ring generated by the irreducible complex representations of $G$ (up to equiv.) for $\oplus$ and $\otimes$. It is a fusion ring, and it is simple iff $G$ is simple. So the notion of simple fusion ring generalizes the notion of simple group; it doesnotcorrespond to the usual notion of simple ring.The fusion ring $\mathcal{G}_p$ is the Grothendieck ring of the cyclic group of prime order $p$.

*Definition*: The fusion ring $\mathcal{F}$ is of multiplicity one if every $ n_{ij}^k \in \{0,1\}$.

*Lemma*: Let $\mathcal{F}$ be a fusion ring of multiplicity one and rank $r$, then FPdim$(\mathcal{F}) \le r^3$.

*Proof*: $d(h_i)^2 = \sum_k n_{ii}^kd(h_k) \le \sum_k d(h_k) \le \sum_k (\sum_s d(h_s))^{1/2} = r (\sum_s d(h_s))^{1/2}$

Let $x = \sum_k d(h_k)>0$. Then, $x^2 \le r^2x$, and so $x \le r^2$. It follows that $d(h_i) \le r$.

But FPdim$(\mathcal{F}) = \sum_i d(h_i)^2 \le \sum_i r^2 = r^3$. $\square$

*Digression*: at multiplicity $m$, we get idem that $d(h_i) \le mr$ and FPdim$(\mathcal{F}) \le m^2r^3$.

*Theorem*: There is no integral simple fusion ring of Frobenius type, multiplicity one and rank $\le 10$ (except $\mathcal{G}_p$).

*Proof*: By the previous lemma, a fusion ring of multiplicity one and rank $\le 10$, has FPdim $ \le 10^3$.

But by a SAGE computation (with this code), there is no integral simple fusion ring of Frobenius type, multiplicity one, rank $\le 10$ and FPdim $ \le 1000$ (except $\mathcal{G}_p$). $\square$

**Question**: Is there an integral simple fusion ring of multiplicity one and Frobenius type (except $\mathcal{G}_p$)?