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.



Non-symmetric Braiding on finite group Representation Categories
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 $S\_3$.


