# R-matrices and symmetric fusion categories

Given a $$\mathbb{C}$$-linear braided fusion category $$\mathcal{C}$$ containing a fusion rule of the form e.g. $$\begin{equation}X\otimes Y\cong A\oplus B \oplus C\end{equation}$$ (where $$A,B, C, X$$ and $$Y$$ are all simple objects with $$A, B, C$$ non-isomorphic), we can write the $$R$$-matrix $$R^{XY}=\text{diag}(R^{XY}_{A}, R^{XY}_{B}, R^{XY}_{C})$$. My intuition has always been that these scalars for two distinct objects cannot be the same (much like eigenvalues for distinct eigenspaces should not be the same). Is this misguided? That is,

(Q) Can we have scalars $$R^{XY}_{A}=R^{XY}_{B}$$ for $$A\not\cong B$$?

If $$\mathcal{C}$$ is symmetric, then I believe that diagonal matrix $$R^{XY}$$ can only have $$\pm1$$s along its diagonal. However, if the answer to (Q) is no, then this would mean that $$\mathcal{C}$$ can only contain fusion rules of the form $$X\otimes Y\cong pA\oplus qB$$ and $$X\otimes Y\cong pA$$ (where $$p$$ and $$q$$ are positive integers). This seems too strong.

Thanks!

Basically any representations of any groups will give you counterexamples in the symmetric case. The simplest case, where X and Y are the defining representation of SU(2), already works. For non-symmetric categories you'd expect that "generically" they'll look different, e.g. if you look at the category of representations of a quantum group and vary q, then for all but finitely many q the eigenvalues will be different. But there's plenty of special q where you'll get repeats, and not just symmetric cases. For example, take SU(2) at a level which has a $$D_{2n}$$ de-equivariantization, then you have a transparent invertible object and if you take the tensor square of the "middle" object it will have both the trivial and the transparent as summands with the same eigenvalue.
• Thanks very much, this is exactly what I was looking for. A follow-up: if e.g. $R^{XY}_{A}$ is now a $k$ by $k$ matrix with $k>1$ (i.e. $A$ appears in the decomposition of $X\otimes Y$ with multiplicity $k$) then suppose we diagonalise this $R$-matrix. Is it also possible for these diagonals to be distinct? – Meths Jun 5 at 10:42
• A partial answer to this for $\mathcal{C}$ ribbon: we know $\left[R^{XY}_{A}R^{YX}_{A}\right]_{\mu\nu}=\frac{\vartheta_{A}}{\vartheta_{X}\vartheta_{Y}}\delta_{\mu\nu}$. In some choice of gauge, we should be able to get $R^{XY}_{A}=R^{YX}_{A}$. For such a choice of gauge, the diagonalisation of $R^{XY}_{A}$ will have scalars $\pm\sqrt{\frac{\vartheta_{A}}{\vartheta_{X}\vartheta_{Y}}}$ along the diagonal. The 'eigenvalues' thus differ by a minus sign at most in this choice of gauge, and if $X=Y$ then the argument is gauge-invariant. – Meths Jun 10 at 13:53