Non-negative integer matrix representation of a fusion ring

Question

Context: I am a physics grad student working on topological lines in 2D CFTs.

Let $A$ be a unital based $\mathbb{Z}_{+}$ ring with finite rank (or a Fusion ring) with the basis $B = \{b_1, b_2, \dotsc b_{n}\}$ such that $$b_i . b_j = \sum_k {N_i^k}_j b_k.$$ The unit element $1 \in B$.

For every $b_i \in B$ there is a unique $b_{i^*} \in B$ such that $$b_i.b_{i^*} = 1 + \sum_{k: b_k \neq1}n^k b_k \quad \text{and} \quad b_{i^*}.b_i = 1 + \sum_{k: b_k \neq1}m^k b_k$$ where $n^k, m^k \in \mathbb{Z}_{\geq 0}$. I will call $b_{i^*}$ as the dual of $b_{i}$ for simplicity.

Consider a $\mathbb{Z}_{+}$ module $M$ over $A$ with the $\mathbb{Z}$ basis $\{m_l\}_{l\in L}$ such that $$b_i m_l = \sum_k a_{il}^k m_k.$$ For each $b_i$ we have a non-negative integer matrix representation i.e. $a_{il}^k$. Is it true that this matrix rep of $b_i$ and its dual $b_{i^*}$ transpose of each other? That is, does it always hold $a_{il}^k = a_{{i^*}k}^{\;l}$? How can we prove this? If not, then is there some other similar relationship between these matrices?

By Frobenius reciprocity, the matrices $N_{il}^k$ and $N_{{i^*}l}^k$ are transpose of each other i.e. $N_{il}^k = N_{{i^*}k}^l$ but I am not sure this property holds for any non-negative intger matrix representation of a fusion ring.

In all the examples, I have seen, this always holds and it feels like it should be true; but I could be wrong.

Answer 1

Here is a standard counterexample. Let $A$ be the ring of rank 2 with basis $1,X$ and $X^2=1+2X$ (so $X=X^*$). Then $A$ has a rank 2 module $M$ where $X$ acts via matrix $\left( \begin{array}{cc}1&1\\2&1\end{array}\right)$ (this matrix satisfies the same equation as $X$ by the Cayley-Hamilton theorem). This is a counterexample since this matrix is not symmetric.

Answer 2

I think I found a counterexample if the characteristic of $M$ is nonzero:

Let $B$ be the matrices corresponding to the permutations of the alternating group of $3$ elements, denoted as $\{ 1,x,x^* \} $ where $x^* = x^2 = x^{-1}$. Define $M$ as the span of the basis $\{1, \omega \}$ modulo $7$ for $\omega = e^{\frac{2 \pi i}{3}}$.

Now do this action of $B$ on $M$: $x$ multiplies with $2 \omega$ and $x^* $ with $3 \omega + 3$ which modulo $7$ is $4 \omega^2$. You can check that this is a module over $A$. And as matrix notation $$a_{x i}^{k} = \begin{pmatrix} 0 & 5\\ 2 & 5\\ \end{pmatrix}$$ while $$a_{x^* i}^{k} = \begin{pmatrix} 3 & 4\\ 3 & 0\\ \end{pmatrix} = \begin{pmatrix} 0 & 5\\ 2 & 5\\ \end{pmatrix}^2$$

Of course this is not the case if you take the more trivial respresentation where $x$ just multiplies with $\omega$, but it's a valid representation.