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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.

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up vote 6 down vote accepted


No, it does not follow.

In this paper (Example 6.14) 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.

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Thank you for answer! I also thought that might not be true in general. – Sebastian Burciu May 30 '10 at 6:56

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