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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). 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):
$$\begin{smallmatrix}1&0&0&0&0&0 \\\ 0&1&0&0&0&0 \\\ 0&0&1&0&0&0 \\\ 0&0&0&1&0&0 \\\ 0&0&0&0&1&0 \\\ 0&0&0&0&0&1\end{smallmatrix} , \ \begin{smallmatrix}0&1&0&0&0&0 \\\ 0&0&1&0&0&0 \\\ 1&0&0&0&0&0 \\\ 0&0&0&0&1&0 \\\ 0&0&0&0&0&1 \\\ 0&0&0&1&0&0\end{smallmatrix} , \ \begin{smallmatrix}0&0&1&0&0&0 \\\ 1&0&0&0&0&0 \\\ 0&1&0&0&0&0 \\\ 0&0&0&0&0&1 \\\ 0&0&0&1&0&0 \\\ 0&0&0&0&1&0\end{smallmatrix} , \ \begin{smallmatrix}0&0&0&1&0&0 \\\ 0&0&0&0&0&1 \\\ 0&0&0&0&1&0 \\\ 1&0&0&1&1&1 \\\ 0&0&1&1&1&1 \\\ 0&1&0&1&1&1\end{smallmatrix} , \ \begin{smallmatrix}0&0&0&0&1&0 \\\ 0&0&0&1&0&0 \\\ 0&0&0&0&0&1 \\\ 0&1&0&1&1&1 \\\ 1&0&0&1&1&1 \\\ 0&0&1&1&1&1\end{smallmatrix} , \ \begin{smallmatrix}0&0&0&0&0&1 \\\ 0&0&0&0&1&0 \\\ 0&0&0&1&0&0 \\\ 0&0&1&1&1&1 \\\ 0&1&0&1&1&1 \\\ 1&0&0&1&1&1\end{smallmatrix}$$

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 $5$?


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.


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.

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  • 2
    $\begingroup$ Dave Penneys, Henry Tucker, and I looked into this several years ago. I don't know if there is anything further on this, but the answer is probably 'All fusion rings of rank 5 are commutative.' We have a draft, but the argument is incomplete. $\endgroup$ – Matthew Titsworth Dec 18 '19 at 2:02
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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 $\ge 5$.

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    $\begingroup$ Very nice! I hadn't quite put together that total positivity and integrality of formal codegrees don't require a categorification to exist. (Aside since it confused me for a minute: a and b are rational integers because Galois conjugation permutes formal codegrees of the same sized matrix blocks. So Galois conjugation can't switch a and b.) $\endgroup$ – Noah Snyder Dec 20 '19 at 18:25

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