I'm trying to understand braidings on finite group representations. They are the same as quasitriangular structures on the group algebra $k[G]$. The original reference seems to be http://arxiv.org/pdf/q-alg/9706007.pdf. It is stated there [sic]:
Teorem 3 Universal $R$-matrix in the group algebra $k[G]$ of a group $G$ is defined by [...] by nondegenerated bimultiplicative $G$-invariant form $\beta: \hat{A} \otimes \hat{A} \to k$.
The formula he gives is: $$R = \frac{1}{\lvert A\rvert} \sum_{\substack{\chi_1, \chi_2 \in \hat{A}.\\g_1,g_2 \in A}}\beta(\chi_1,\chi_2) \chi_1(g_1) \chi_2(g_2) \cdot g_1 \otimes g_2$$
And then:
Unitary $R$-matrix corresponds to the case of [...] skewsymmetric form.
"Unitary" here means $RR_{21} = 1 \otimes 1$, also known as "triangular", i.e. it gives a symmetric braiding.
Later in the proof, it only says:
The direct cheking shows the equivalence of this condition to skewsymmetricity of the corresponding form.
I can't see at all why a skewsymmetric $\beta$ makes $RR_{21} = 1 \otimes 1$ true. This is how far I get:
$$RR_{21} = \frac{1}{\lvert A\rvert^2} \sum_{\substack{\chi_i, \in \hat{A}\\g_i \in A\\i \in \{1...4\}}} \beta(\chi_1,\chi_2) \beta(\chi_3,\chi_4) \left(\prod_i \chi_i(g_i)\right) \cdot g_1g_4 \otimes g_2g_3$$
And now I'm supposed to insert $\beta(\chi_1,\chi_2) = \beta(\chi_2,\chi_1)^{-1}$ and suddenly all summands vanish except $1 \otimes 1$? It's just not happening.
Or maybe I'm not understanding what "skewsymmetric" is supposed to mean here? There is an earlier bit where he says that the Markov element $u = S(R_{(1)})R_{(2)}$ satisfies $\chi(u) = \beta(\chi,\chi)$, but isn't the right hand side automatically 1?
Edit: I'm learning that the following is probably relevant: There is a quantity $Q = R R_{21}$ and a map $\Phi: kG \to k[G]$ given by $\Phi(f) = \sum f(Q_{(1)})Q_{(2)}$. If this map is an isomorphism, the braiding makes $\operatorname{Rep}(G)$ into a modular category, if it's "trivial", the braiding is symmetric. But I don't see what trivial exactly means here.