Suppose that $A$ is a semisimple Hopf algebra with a commutative character ring. Does it follow that $A$ is quasitriangular, i.e $\mathrm{Rep}(A)$ is a braided tensor category?


I think I 've seen this statement in a paper without a proof long time ago. It might be obvious although I don't see how to construct a braiding just knowing non-functorial commutativity of the tensor products.



Semisimple Hopf algebras with commutative character ring
Sebastian,


No, it does not follow.


In [this paper (Example 6.14)](http://lanl.arxiv.org/abs/0704.0195) we proved that if a Tambara-Yamagami fusion category 
admits a braiding then its dimension is a power of 2. Note that a Tambara-Yamagami
category
has a commutative Grothendieck ring. Hopf algebras whose representation
category is of Tambara-Yamagami type are classified by Tambara (Representations of tensor categories with fusion rules of self-duality for abelian groups, Isr. J. Math. 118 (2000), 29-60). For example, there is a Hopf algebra $A = k^9 \oplus M\_3(k)$ (so-called Kac-Paljutkin algebra) with commutative chracater ring and $Rep(A)$ admitting no braiding. 


