Do the fusion categories $Rep(S_4)$ and $Rep(A_5)$ admit non-symmetric braidings? All the other rep. cats. of finite subgroups of $SU(2)$ do (in the McKay correspondence). My guess is no.
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asked May 28, 2010 at 14:38
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2 Answers
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Eric,
The answer is no for $Rep(A_5)$ and yes for $Rep(S_4)$, thanks to Victor Ostrik's observation. For a braided category $C$ let $C'$ denote its Mueger center, i.e., the subcategory of objects $Y$ in $C$ such that the square of braiding of $Y$ with any $X$ in $C$ is identity. So $C$ is symmetric if $C=C'$ and $C$ is non-degenerate (or modular) if $C'$ is trivial.
Note that $C:=Rep(A_5)$ is simple, i.e., it has no non-trivial proper fusion subcategories. Now if $C$ has a non-symmetric braiding then $C' \neq C$ is a proper subcategory. So $C'$ is trivial, i.e., $C$ with the above braiding is non-degenerate (modular). This cannot happen (e.g., $C$ has a simple object of dimension 5, but in a modular category the square of dimension of any object divides dimension of the category, thanks to the result of Etingof-Gelaki).
For $D:= Rep(S_4)$ there is a non-symmetric braiding with $D'=Rep(S_3)$, namely the equivariantization of a pointed category $Vec_{Z/2Z\oplus Z/2Z}$ with respect to an action of $S_3$.
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1$\begingroup$ Thanks Dimitri! How do you rule out $D^\prime=Rep(S_3)$? $\endgroup$ May 28, 2010 at 16:33
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3$\begingroup$ Dmitri, for $D=Rep(S_4)$ there is one more possibility, namely $D'=Rep(S_3)$. Actually I believe that for $D=Rep(S_4)$ the alternative braiding is possible. Namely consider pointed category with underlying group $Z/2Z \oplus Z/2Z$ and quadratic form which takes value -1 on a nontrivial element. Then equivariantize with respect to the obvious $S_3-$action. $\endgroup$ May 28, 2010 at 16:40
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$\begingroup$ Victor is right. I will edit the above answer. $\endgroup$ May 28, 2010 at 16:49
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You can also answer the question using the classification of R-matrices over group algebras. Indeed, the R-matrices in the Hopf algebra $ kG $, are classified by pairs $ (A, \rho) $ where $ A \subset G $ is an abelian normal subgroup and $ \rho: A \times A \to k^* $ is a bilinear form ad $ G $-invariant. It is easy to see that the associated R-matrix is symmetric if and only if the bilinear form is skew-symmetric.
It is now clear that $ kA_5 $ does not have R-matrices in addition to the trivial. On the other hand, the only normal abelian subgroup of $ S_4 $ is formed by the union of the conjugacy classes of $(12)(34)$ and the identity of $S_4$. This normal subgroup is isomorphic to $C_2 \times C_2 $. As was mentioned by Ostrik, one example could be the next: the quadratic form that takes values -1 on nontrivial elements. That bilinear form satisfies the conditions and defines a non-symmetric R-matrix.
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$\begingroup$ Thanks Cesar. What is a reference for this classification? Is there a variation for twisted group algebras? So this might lead to a more general statement for categories with the same fusion rules as $Rep(G)$ for $G$ a non-abelian simple group? $\endgroup$ Jun 3, 2010 at 13:30
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$\begingroup$ 1) You can see the result in Example 2.1 of arXiv: math/0605731, or arXiv: q-alg/9706007. 2) On the variation for the twisted group algebras, I do not know anything, but I do not know which would be a possible variation. 3) The result implies that the only braid on Rep (G) for G a non-abelian simple group is the trivial braid. But I think you can not say anything specific for fusion categories with the same fusion rules of Rep(G), since there are pairs of fusion categories with the same rules of fusion, where one of them is braided and the other has not a braid. $\endgroup$ Jun 3, 2010 at 15:03
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$\begingroup$ Cesar, I was thinking of using Prop. 9.11 of 0809.3031 to say something about when the only braidings on a category are symmetric. Thanks for the references. $\endgroup$ Jun 3, 2010 at 18:30