A fusion ring $\mathcal{F}$ (of rank $r$) 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](https://mathoverflow.net/q/344079/34538)). The fusion ring $\mathcal{F}$ is called *noncommutative* if $\exists i,j$ with $b\_ib\_j\neq b\_jb\_i$. 


**Question**: What is the smallest rank for a noncommutative fusion ring? 


We already know that this smallest rank is at most $6$ because the Grothendieck ring of the Haagerup fusion category $H\_6$ is noncommutative and of rank $6$. Here are its fusion rules (coming from [this paper](https://mathscinet.ams.org/mathscinet-getitem?mr=2909758)):  

$$\begin{smallmatrix}1&0&0&0&0&0 \\\ 0&1&0&0&0&0 \\\ 0&0&1&0&0&0 \\\ 0&0&0&1&0&0 \\\ 0&0&0&0&1&0 \\\ 0&0&0&0&0&1\end{smallmatrix} , \
\begin{smallmatrix}0&1&0&0&0&0 \\\ 0&0&1&0&0&0 \\\ 1&0&0&0&0&0 \\\ 0&0&0&0&1&0 \\\ 0&0&0&0&0&1 \\\ 0&0&0&1&0&0\end{smallmatrix} , \
\begin{smallmatrix}0&0&1&0&0&0 \\\ 1&0&0&0&0&0 \\\ 0&1&0&0&0&0 \\\ 0&0&0&0&0&1 \\\ 0&0&0&1&0&0 \\\ 0&0&0&0&1&0\end{smallmatrix} , \
\begin{smallmatrix}0&0&0&1&0&0 \\\ 0&0&0&0&0&1 \\\ 0&0&0&0&1&0 \\\ 1&0&0&1&1&1 \\\ 0&0&1&1&1&1 \\\ 0&1&0&1&1&1\end{smallmatrix} , \
\begin{smallmatrix}0&0&0&0&1&0 \\\ 0&0&0&1&0&0 \\\ 0&0&0&0&0&1 \\\ 0&1&0&1&1&1 \\\ 1&0&0&1&1&1 \\\ 0&0&1&1&1&1\end{smallmatrix} , \
\begin{smallmatrix}0&0&0&0&0&1 \\\ 0&0&0&0&1&0 \\\ 0&0&0&1&0&0 \\\ 0&0&1&1&1&1 \\\ 0&1&0&1&1&1 \\\ 1&0&0&1&1&1\end{smallmatrix}$$ 


We also know that this smallest rank is at least $5$, because $\mathbb{C}B$ admits a structure of finite dimensional von Neumann algebra of the form $\mathbb{C} \oplus A$ (where the first component is generated by $\sum\_id(b\_i)b\_i$). By noncommutativity, if $A$ has the smallest possible dimension then $A = M\_2(\mathbb{C})$. Then a noncommutative fusion ring is of rank at least $5$. So the question reformulates as follows: 


**Reformulated question**: Is there a noncommutative fusion ring of rank $5$? 




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*Investigation at rank $5$* 


First note that $ n\_{j^\*,i^\*}^{k^\*} = n\_{i,j}^k $ because $b\_{j^\*}b\_{i^\*} = (b\_i b\_j)^\* = \sum\_k n\_{i,j}^k b\_{k^\*}$, so if $i^\* = i$ for all $i$ then the fusion ring is commutative. Thus there is $i$ such that $i^\* \neq i$.  

We can assume that $2^\* = 3$. Then there are two cases:  

(1) $4^\* = 5$,  

(2) $4^\*=4$ (and so $5^\* = 5$). 


Note that (1) implies commutativity (see Appendix below). So we can assume (2), and then by Frobenius reciprocity, the fusion rules must be as follows (with $16$ parameters):  

$$\begin{smallmatrix}1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\end{smallmatrix}, \ 
\begin{smallmatrix}0&1&0&0&0\\0&a\_1&a\_{11}&a\_6&a\_{10}\\1&a\_1&a\_1&a\_2&a\_3\\0&a\_4&a\_6&a\_7&a\_8\\0&a\_5&a\_{10}&a\_9&a\_{12}\end{smallmatrix} , \ 
\begin{smallmatrix}0&0&1&0&0\\1&a\_1&a\_1&a\_4&a\_5\\0&a\_{11}&a\_1&a\_6&a\_{10}\\0&a\_6&a\_2&a\_7&a\_9\\0&a\_{10}&a\_3&a\_8&a\_{12}\end{smallmatrix} , \ 
\begin{smallmatrix}0&0&0&1&0\\0&a\_2&a\_6&a\_7&a\_9\\0&a\_6&a\_4&a\_7&a\_8\\1&a\_7&a\_7&a\_{13}&a\_{15}\\0&a\_9&a\_8&a\_{15}&a\_{16}\end{smallmatrix} , \ 
\begin{smallmatrix}0&0&0&0&1\\0&a\_3&a\_{10}&a\_8&a\_{12}\\0&a\_{10}&a\_5&a\_9&a\_{12}\\0&a\_8&a\_9&a\_{15}&a\_{16}\\1&a\_{12}&a\_{12}&a\_{16}&a\_{14}\end{smallmatrix}$$
such that $a\_i \in \mathbb{Z}\_{\ge 0}$, and $\sum\_s n\_{i,j}^sn\_{s,k}^t = \sum\_s n\_{j,k}^sn\_{i,s}^t$ (associativity). Note that by brute-force computation, there is no noncommutative example of multiplicity at most three in this case.




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*Appendix*: the proof that (1) implies commutativity. 


$$b\_2 b\_3 = b\_1 + n\_{2,3}^2b\_2 + n\_{2,3}^3b\_3 + n\_{2,3}^4b\_4 + n\_{2,3}^5b\_5,$$ 
but $(b\_2b\_3)^\* = b\_3^\* b\_2^\* = b\_2b\_3$, so $n\_{2,3}^2 = n\_{2,3}^3$ and $n\_{2,3}^4=n\_{2,3}^5$. It follows that $$b\_2 b\_3 = b\_1 + n\_{2,3}^2(b\_2 + b\_3) + n\_{2,3}^4(b\_4 + b\_5).$$ Idem, $b\_3 b\_2 = b\_1 + n\_{3,2}^3(b\_2 + b\_3) + n\_{3,2}^4(b\_4 + b\_5)$. By Frobenius reciprocity, $n\_{2,3}^2 = n\_{3,2}^3$, but $d(b\_2b\_3) = d(b\_3b\_2)$, so $n\_{2,3}^4d(b\_4 + b\_5) = n\_{3,2}^4d(b\_4 + b\_5)$. Then $n\_{2,3}^4 = n\_{3,2}^4$ and $b\_2b\_3 = b\_3b\_2$. Idem $b\_4b\_5 = b\_5b\_4$. Next by Frobenius reciprocity: 


* $n\_{2,4}^2 = n\_{3,2}^4 = n\_{2,3}^4 = n\_{4,2}^2$ (the second equality comes from $b\_2b\_3 = b\_3b\_2$),
* $n\_{2,4}^3 = n\_{3,3}^4 = n\_{4,2}^3$,
* $n\_{2,4}^4 = n\_{4,5}^2 = n\_{5,4}^2 = n\_{4,2}^4$ (the second equality comes from $b\_4b\_5 = b\_5b\_4$),
* $n\_{2,4}^5 = n\_{5,5}^2 = n\_{4,2}^5$.


It follows that $b\_2b\_4=b\_4b\_2$. Idem $b\_2b\_5=b\_5b\_2$, $b\_3b\_4=b\_4b\_3$ and $b\_3b\_5=b\_5b\_3$. 


*Conclusion*: the fusion ring is commutative in this case. 



What is the smallest rank for a noncommutative fusion ring?
I think that a noncommutative fusion ring of rank 5 does not exist. Namely, let $a$ and $b$ be the formal codegrees (see <https://arxiv.org/pdf/0810.3242.pdf>) of such ring. Then $a$ and $b$ are positive (EDIT: and rational, see the explanation by Noah) integers satisfying $\frac1a+\frac2b=1$ (see Proposition 2.10 in <https://arxiv.org/pdf/1309.4822.pdf>). It is easy to see that this equation has no solutions with $a\ge 5$. This is impossible as the Frobenius-Perron dimension should be $\ge 5$.


