Let $\mathbb{A}$ be the ring of algebraic integers. Consider a sequence $(d_i)_{i \in I}$, with $I$ a finite set and $d_i \in \mathbb{A} \cap \mathbb{R}_{\ge 1}$, such that $$d_i d_j = \sum_{k \in I} n_{i,j,k} d_k,$$ for all $i,j \in I$, with $n_{i,j,k} \in \mathbb{Z}_{\ge 0}$.

A subset $J \subset I$ is called a subsystem if $\forall i,j \in J$ and $\forall k \not \in J$ then $n_{i,j,k} = 0$. Let $\langle i \rangle$ be the smallest subsystem containing $i$.

Question: Let $i,j \in I$ such that $\langle i \rangle = \langle j \rangle$. Is it true that $\mathbb{Z}[d_i] = \mathbb{Z}[d_j]$?

Observation: The vector $v = (d_k)_{k \in I}$ is a common eigenvector for the matrices $M_i = (n_{i,j,k})_{j,k \in I}$, with eigenvalue $d_i$, because: $$ M_i v = (\sum_{k \in I} n_{i,j,k} d_k)_{j \in I} = (d_i d_j)_{j \in I} = d_i v.$$ By Frobenius-Perron Theorem (stated below), the eigenvalue $d_i$ of $M_i$ is its spectral radius.

Frobenius-Perron Theorem: A square matrix $M$ with nonnegative real entries has a non-negative real eigenvalue. If moreover $M$ has an eigenvector $v$ with strictly positive entries, then the eigenvalue of $v$ is the largest non-negative real eigenvalue and is the spectral radius of $M$.


1 Answer 1


No, here are infinitely many counter-examples for $I = \{1,2\}$.

Let $n,m \in \mathbb{Z}_{\ge 1}$ such that $m | n^2$, $n | 2m^2$ and $n \neq 2m$ (e.g. $n=m=1$).

Take $d_1 = n(1+\sqrt{2})$, $d_2 = m(3+2\sqrt{2})$. Here are the matrices $M_i = (n_{i,j,k})_{j,k \in I}$:

$$\left(\begin{matrix}0&\frac{n^2}{m}\\m&2n \end{matrix}\right), \ \left(\begin{matrix}m&2n\\\frac{2m^2}{n}&5m\end{matrix}\right)$$

Now $\langle 1 \rangle = \langle 2 \rangle = I$, but $\mathbb{Z}[d_1] = \mathbb{Z}[n \sqrt{2}] \neq \mathbb{Z}[2m \sqrt{2}] = \mathbb{Z}[d_2]$, because $n \neq 2m$.

Here is a counter-example which is also a fusion ring (the initial motivation), with $I = \{1,2,3\}$.

Take $d_1 = 1$, $d_2 = 3+2\sqrt{2}$, $d_3 = 4+3\sqrt{2}$. Here are the matrices $(M_i)$:

$$\left(\begin{matrix}1&0&0\\0&1&0\\0&0&1 \end{matrix}\right), \ \left(\begin{matrix}0&1&0\\1&0&4\\0&4&3 \end{matrix}\right), \ \left(\begin{matrix}0&0&1\\0&4&3\\1&3&6 \end{matrix}\right)$$

Now $\langle 2 \rangle = \langle 3 \rangle = I$, but $\mathbb{Z}[d_2] = \mathbb{Z}[2\sqrt{2}] \neq \mathbb{Z}[3\sqrt{2}] = \mathbb{Z}[d_3]$.

For the notion of fusion ring, see Definition 3.1.7 in the following reference:
P. Etingof, S. Gelaki, D. Nikshych, V. Ostrik; Tensor categories. Mathematical Surveys and Monographs (2015) 205.

Conclusion: We should replace $\mathbb{Z}[d_i] = \mathbb{Z}[d_j]$ by $\mathbb{Q}(d_i) = \mathbb{Q}(d_j)$ in the question.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.