# Is there an integral simple fusion ring of multiplicity one and Frobenius type? (obvious excepted)

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

A fusion ring is 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 called integral if every $d(h_i)$ is an integer.
Its Frobenius-Perron dimension (FPdim) is $\sum d(h_i)^2$.
It is of Frobenius type if every $d(h_i)$ divides FPdim$(\mathbb{C}\mathcal{B})$
It is simple if $r>1$ 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 does not correspond 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 $\mathcal{G}_p$)?

• What about Fibonacci or Tambara-Yamagami? – Dave Penneys Jul 8 '16 at 6:27
• @DavePenneys: Fibonacci is not integral and Tambara-Yamagami is not simple. – Sebastien Palcoux Jul 8 '16 at 9:44
• Specifically, Tambara-Yamagami are $\mathbb Z_2$ graded. A simple fusion ring would have to be trivially graded. – Matthew Titsworth Jul 8 '16 at 14:55
• Yeah, somehow I missed both the words "integral" and "simple" and was merely thinking about multiplicity one... Sorry about that. – Dave Penneys Jul 10 '16 at 16:16