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Consider a commutative ring $x_ix_j = N_{ij}^k x_k$, where $N_{ij}^k \in\{0,1,2,3,\cdots\}$, and $\{x_i\}$ is a finite set. (This is actually a fusion ring and $x_i$ are simple objects.)

How to find the irreducible integer-matrix representations of the above ring? (ie what is the algorithm to find the irreducible representations given $N_{ij}^k$.)

(Given a group multiplication table of a finite group, there is a way to construct a (reducible) canonical integer-matrix representations by considering a vector space whose basis vectors are labelled by the group elements. Since such reducible representation contain all the irreducible representations, so we can try to decompose the canonical representations to find all the irreducible representations. Do we have similar results for integer-matrix representation of rings?)

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    $\begingroup$ Do you really want modules over this ring? I would guess that you would instead want module categories over the corresponding fusion category (assuming there is one). $\endgroup$
    – Qiaochu Yuan
    Commented Nov 17, 2015 at 17:38
  • $\begingroup$ Yes, I really want modules over this ring. Although I am also interested in module categories over the corresponding fusion category, but I do not have a list of fusion categories. I do have a list of fusion rings. So I like to start with fusion rings. Note that I asked for an algorithm to find the irreducible representations from the integer data $N^k_{ij}$. $\endgroup$
    – Xiao-Gang Wen
    Commented Nov 17, 2015 at 17:44
  • $\begingroup$ Hi Xiao-Gang, I don't have a full answer, but here is my thinking: Associativity of the ring ensure the matrix $\rm N_i$ itself is a representation. Question is whether it is reducible. Now if instead we work over field $k$, i.e., release the constrain of integer-matrix rep. According to Artin–Wedderburn theorem, the algebra here should be a direct sum of matrix algebra, and due to the commutativity you provide, these matrix algebra should commutes and therefore 1-dimensional. So as an algebra over field $k$, it is reducible, and its 1-dimensional rep can be achieved by diagonalizing $\endgroup$
    – Yingfei Gu
    Commented Dec 10, 2015 at 20:24
  • $\begingroup$ matrix $N_i$, and each eigenvalue produces a rep, so there are precisely $r=rank$ many 1-dimensional rep you can immediately get from data $N$. The next task is to figure out whether there is a way to put a subset, say pick $t<r$, of these 1-dimensional reps (most possibly some complex numbers) make diagonal matrices $N_i'=diag(\lambda_i^{(1)},\ldots,\lambda_i^{(t)})$, and find a basis transformation back to an integer matrix. $\endgroup$
    – Yingfei Gu
    Commented Dec 10, 2015 at 20:31
  • $\begingroup$ This is certainly difficult, but in practical, it is possible to rule out many possibilities. For example, the matrix $N_i'$ have the same determinant of an integer matrix, therefore, $\prod_{j=1}^t \lambda_i^{(j)} \in \mathbb{Z}$ for all $i$. Maybe with other knowledge of integer matrix we can reduce the solutions further. $\endgroup$
    – Yingfei Gu
    Commented Dec 10, 2015 at 20:37

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