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It's a rather obvious idea in the area of fusion rings, but I haven't found a reference yet. Start with the usual rules for a rank n fusion ring
$X_i\bigotimes{X_j}=\Sigma_k{T_{ij}}^kX_k$
and interpret the objects $X_i$ as n diagonal matrices $D_{ii}$ and $\bigotimes$ as ordinary matrix multiplication. The (more or less unique) solution of this system is even simpler: Interpret the $X_i$ as scalars and $\bigotimes$ as very ordinary :-) multiplication. $D_{ii}$ is then the i-th solution of this system (having exactly n solutions), placed on the diagonal. You can collect the n diagonals in a final n*n matrix $M_{ij}$. (Some freedom of row/column permutation - make sure that the first row consists only of 1, put the other rows in any order.)
You now can combine $M$ and the Verlinde $S$ matrix to a lot of "cool" equations, e.g. (+ is transpose) $A=MS=(MS)^+$ (A is symmetric) or $B=S^{+-}M$ (B is diagonal). Computing $S$ from $M$ is very easy (even if $S$ is NOT symmetric!).
I called the $D_{ii}$ "generators" because this all resembles, especially in graphic form (Dinotracks - like Birdtracks, only different :-) somewhat Lie algebra generators (you also have a "Jacobi relation" etc.).
Do you know a paper where the matrices M have been put to good use? (I have no idea if computing the Verlinde $S$ matrix from given fusion rules is considered as a "hard" problem, and the feeling that classification of fusion rings works the other way round anyway - restrict $S$ and compute all possible $T$ that remain.)
Can you give a proof for the "cool" equations? (I merely observed them.)
Here are some more if you are interested: http://imgur.com/1ZfBVTm
(Dropping arrows - matrix is symmetric and real, hole in dot - it is also diagonal)

Bonus actual example:
Rule: $A\bigotimes{A}=1,A\bigotimes{B}=B,B\bigotimes{B}=1\bigoplus{A}\bigoplus{B}$
$M=\begin{pmatrix} 1 & 1 & 1\\ 1 & 1 & -1\\ 2 & -1 & 0 \end{pmatrix}$
$S=\begin{pmatrix} 1/\surd{6} & 1/\surd{6} & \surd{2/3}\\ 1/\surd{3} & 1/\surd{3} & -1/\surd{3}\\ 1/\surd{2} & -1/\surd{2} & 0 \end{pmatrix}$
(Any chemist will observe M is the character table of $C_{3v}$ and the multiplication table is the same but this is only half an accident :-)

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  • $\begingroup$ The fusion ring you are prescribing is the even part of the subfactor $R^G\subset (R\otimes M_2)^G$ with index 4, where $G\subset \mathrm{SU}(2)$ is the group associated with the affine Dynkin dynkin diagram $D^{(1)}_5$, I guess it is the binary dihedral group $BD_5=Q_3$. $\endgroup$ – Marcel Bischoff Jan 2 '15 at 18:35
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This is just an answer to part of the question, namely how to determine $S$ from the fusion rules in the case of "modular data". See also chapter 5, here http://www.theorie.physik.uni-goe.de/papers/rehren/89/braid_group_statistics.pdf for a comparison between the characters of a group and the $S$-matrix of a unitary modular tensor category.

If you know all fusion matrices, you can in principle calculate the Verlinde matrix $S$. You need that the fusion rules are abelian. Then the matrices $N_i=(T^k_{ij})$ are commuting and you find a common set of eigenvectors $e_i$, i.e. $$ N_j e_i=\lambda_{j,i} e_i.$$ Let $e_0$ be the vector with the greatest eigenvalue and normalize it such that $e_0=(1,d_1,\ldots,d_n)$. They also fulfill (using Frobenius reciprocity and commutativity) $$ N_iN_j=\sum_k T_{ij}^k N_k. $$ Applying this to $e_0$ you get $$ d_id_j=\sum_k T^k_{ij} d_k, $$ in other words $d_i$ are the Perron-Frobenius dimensions.

Then the Verlinde matrix should be given as: $$ S= \frac1{\sqrt{\sum_i d_i^2}}\left(e_0,\frac{e_{\sigma(1)}}{d_1},\ldots,\frac{ e_{\sigma(n)}}{d_n}\right), $$ where $\sigma$ is a permutation of ${1,\ldots,n}$ and $e_i$ are normalized such that the first entry is 1.

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