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A fusion ring $\mathcal{F}$ is given by a finite set $B = \{b_1,b_2, \dots, b_r \}$ such that $b_i b_j = \sum_k n_{i,j}^k b_k$ with $n_{i,j}^k \in \mathbb{Z}_{\ge 0}$, satisfying axioms slightly augmenting the group axioms (see the details here).

The fusion ring $\mathcal{F}$ is called noncommutative if $\exists i,j$ with $b_ib_j\neq b_jb_i$. A fusion subring is given by a subset $B_S=\{b_s \ | \ s \in S\} \subseteq B$ with $1 \in S = S^*$ and $\forall i,j \in S$ then $n_{i,j}^k \neq 0$ only if $k \in S$. It is called simple when every fusion subring is given by the subset $\{b_1\}$ or $B$. Let $G$ be a finite group, then the Grothendieck ring of $Rep(G)$ is simple (as fusion ring) if and only if $G$ is simple.

Question: Is there a noncommutative simple fusion ring?


Investigation

According to this post, a noncommutative fusion ring must be of rank at least $5$, and at rank exactly $5$, we can assume: $2^*=3$, $4^*=4$ and $5^*=5$. A brute-force computation shows that at multiplicity at most three, there is a unique simple example (upto equiv.):
$$\begin{smallmatrix}1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\end{smallmatrix} , \ \begin{smallmatrix}0&1&0&0&0\\0&0&1&1&0\\1&0&0&0&1\\0&0&1&0&1\\0&1&0&1&1\end{smallmatrix} , \ \begin{smallmatrix}0&0&1&0&0\\1&0&0&0&1\\0&1&0&1&0\\0&1&0&0&1\\0&0&1&1&1\end{smallmatrix} , \ \begin{smallmatrix}0&0&0&1&0\\0&0&1&0&1\\0&1&0&0&1\\1&0&0&1&1\\0&1&1&1&1\end{smallmatrix} , \ \begin{smallmatrix}0&0&0&0&1\\0&1&0&1&1\\0&0&1&1&1\\0&1&1&1&1\\1&1&1&1&2\end{smallmatrix}$$ but it is commutative. Note that $[d(b_1),d(b_2),d(b_3),d(b_4),d(b_5)] =[1,\alpha,\alpha,\beta,\gamma]$ with:

  • $\alpha =1+2cos(2\pi/7) \simeq 2.246979603\dots$,
  • $\beta =1-2cos(6\pi/7) \simeq 2.801937735\dots$,
  • $\gamma =\alpha+\beta-1 \simeq 4.048917339\dots$,
  • so that: $gdim:=\sum_i d(b_i)^2 \simeq 36.65039990\dots$,

moreover, it is of Frobenius type and satisfies Schur Product Property.

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The fusion ring of one of the even parts of Extended Haagerup will work (see Appendix A of our paper). It has rank 8, and you can see that $AB \neq BA$. Surely there's less complicated examples though.

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  • $\begingroup$ Do you expect the existence of an integral example? $\endgroup$ – Sebastien Palcoux Dec 16 '19 at 3:41
  • $\begingroup$ I’d expect there would be integral ones, my feeling is that once the multiplicities are allowed to be large there should be lots of fusion rings. $\endgroup$ – Noah Snyder Dec 16 '19 at 4:24
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Here are the explicit fusion rules for the Extended Haagerup fusion category EH2, mentioned by Noah.

$$\begin{smallmatrix}1&0&0&0&0&0&0&0 \\\ 0&1&0&0&0&0&0&0 \\\ 0&0&1&0&0&0&0&0 \\\ 0&0&0&1&0&0&0&0 \\\ 0&0&0&0&1&0&0&0 \\\ 0&0&0&0&0&1&0&0 \\\ 0&0&0&0&0&0&1&0 \\\ 0&0&0&0&0&0&0&1\end{smallmatrix} , \ \begin{smallmatrix}0&1&0&0&0&0&0&0 \\\ 1&1&1&0&0&0&0&0 \\\ 0&1&1&1&0&0&0&0 \\\ 0&0&1&1&1&1&0&0 \\\ 0&0&0&1&1&1&0&1 \\\ 0&0&0&1&1&1&1&0 \\\ 0&0&0&0&0&1&0&0 \\\ 0&0&0&0&1&0&0&0\end{smallmatrix} , \ \begin{smallmatrix}0&0&1&0&0&0&0&0 \\\ 0&1&1&1&0&0&0&0 \\\ 1&1&1&1&1&1&0&0 \\\ 0&1&1&2&2&2&1&1 \\\ 0&0&1&2&2&2&1&0 \\\ 0&0&1&2&2&2&0&1 \\\ 0&0&0&1&1&0&0&0 \\\ 0&0&0&1&0&1&0&0\end{smallmatrix} , \ \begin{smallmatrix}0&0&0&1&0&0&0&0 \\\ 0&0&1&1&1&1&0&0 \\\ 0&1&1&2&2&2&1&1 \\\ 1&1&2&4&4&4&1&1 \\\ 0&1&2&4&3&4&1&1 \\\ 0&1&2&4&4&3&1&1 \\\ 0&0&1&1&1&1&0&1 \\\ 0&0&1&1&1&1&1&0\end{smallmatrix} , \ \begin{smallmatrix}0&0&0&0&1&0&0&0 \\\ 0&0&0&1&1&1&1&0 \\\ 0&0&1&2&2&2&0&1 \\\ 0&1&2&4&3&4&1&1 \\\ 1&1&2&3&4&3&1&1 \\\ 0&1&2&4&3&3&1&1 \\\ 0&1&0&1&1&1&1&0 \\\ 0&0&1&1&1&1&0&0\end{smallmatrix} , \ \begin{smallmatrix}0&0&0&0&0&1&0&0 \\\ 0&0&0&1&1&1&0&1 \\\ 0&0&1&2&2&2&1&0 \\\ 0&1&2&4&4&3&1&1 \\\ 0&1&2&4&3&3&1&1 \\\ 1&1&2&3&3&4&1&1 \\\ 0&0&1&1&1&1&0&0 \\\ 0&1&0&1&1&1&0&1\end{smallmatrix} , \ \begin{smallmatrix}0&0&0&0&0&0&1&0 \\\ 0&0&0&0&1&0&0&0 \\\ 0&0&0&1&0&1&0&0 \\\ 0&0&1&1&1&1&0&1 \\\ 0&0&1&1&1&1&0&0 \\\ 0&1&0&1&1&1&1&0 \\\ 0&0&0&1&0&0&0&0 \\\ 1&0&0&0&1&0&0&0\end{smallmatrix} , \ \begin{smallmatrix}0&0&0&0&0&0&0&1 \\\ 0&0&0&0&0&1&0&0 \\\ 0&0&0&1&1&0&0&0 \\\ 0&0&1&1&1&1&1&0 \\\ 0&1&0&1&1&1&0&1 \\\ 0&0&1&1&1&1&0&0 \\\ 1&0&0&0&0&1&0&0 \\\ 0&0&0&1&0&0&0&0\end{smallmatrix} $$

It is a rank $8$ noncommutative simple fusion ring. If $\mathcal{B}$ is its basis then $\mathbb{C}\mathcal{B}=\mathbb{C}^4 \oplus M_2(\mathbb{C})$.

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