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
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 :-)

  • $\begingroup$ 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

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.


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.