# What is the smallest rank for a noncommutative fusion ring?

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.

• 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. – Matthew Titsworth Dec 18 '19 at 2:02

## 1 Answer

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 $$\ge 5$$.

• 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.) – Noah Snyder Dec 20 '19 at 18:25