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What is the smallest group $G$ such that $\operatorname{Rep}(G)$ has a non-symmetric braiding (or just an easy example)?

I seem to remember a result classifying all universal $R$-matrices of $\mathbb C[G]$ for finite groups $G$, but cannot find this. But I remember that with this result, I found that the only non-trivial universal $R$-matrix of $C_2$ was

$$R=\frac12(e\otimes e+e\otimes g+g\otimes e-g\otimes g).$$

But this $R$-matrix gives a symmetric braiding on $\operatorname{Rep}(C_2)$. I'm looking for one that will give me a non-symmetric braiding.

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    $\begingroup$ Obvious comment: Rep(C_2) has a non-symmetric braiding giving (a category equivalent to) super vector spaces. I guess you are after something more sophisticated? $\endgroup$
    – Geordie Williamson
    Commented Jun 25, 2023 at 23:06
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    $\begingroup$ @GeordieWilliamson By a braiding $c$ being symmetric, I mean that $c_{W,V}\circ c_{V,W}=\operatorname{id}_{V\otimes W}$. This is the case for the braiding on the category of super vector spaces that you mentioned, and it is in fact the braiding corresponding to the universal $R$-matrix that I mention in my question. $\endgroup$
    – shin chan
    Commented Jun 26, 2023 at 2:35
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    $\begingroup$ You might want to look at the answers to mathoverflow.net/questions/26268. $\endgroup$
    – Dave Benson
    Commented Jun 26, 2023 at 8:15
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    $\begingroup$ @shinchan: of course you're right. Sorry for silly comment. Please ignore. $\endgroup$
    – Geordie Williamson
    Commented Jun 26, 2023 at 8:31
  • $\begingroup$ What is meant by "smallest"? is $G$ meant to be a finite group? $\endgroup$
    – YCor
    Commented Aug 6, 2023 at 8:19

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Since you mention classification results for $R$-matrices:
For finite abelian groups, there is a bijection between the set of universal $R$-matrices of the group hopf algebra $\mathbb C[G]$, the set of bicharacters $\theta: G\times G\rightarrow\mathbb C^*$ on the group $G$ and the set of the braidings of the monoidal Category of representations of the group hopf algebra $\mathbb C[G]$. (If I remember correctly, this was first obtained by Scheunert at Universal R-matrices for finite Abelian groups - a new look at graded multilinear algebra).
This bijection is such that the skew-symmetric bicharacters correspond to triangular $R$-matrices and to symmetric braidings.

So, if you are looking for non-symmetric braidings this is equivalent to looking for non-triangular $R$-matrices or for non skew-symmetric bicharacters on the group. In this sense, an "easy" example can be obtained if you take a symmetric bicharacter of the Klein 4-group $\mathbb Z_2 \oplus\mathbb Z_2$. Plugging this into the formulae given in Sceunert's paper you can obtain (with straightforward computations) the corresponding non-triangular $R$-matrix and the corresponding non-symmetric braiding for the category of the reps of the group hopf algebra.

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    $\begingroup$ For my own sake, I'll try to explicitly write down the above. The non-symmetric $R$-matrix on $G=\mathbf{Z}/2\times\mathbf{Z}/2$ sends $R(a,b)=(-1)^{a_2\cdot b_1}$ where $a=(a_1,a_2)$. This satisfies the Yang-Baxter equation because $R(a+a',b)=R(a,b)R(a',b)$ and likewise for $b+b'$, and almost cocommutativity is automatic because $\mathbf{C}[G]$ is co/commutative. Similarly, given a symmetric $R$-matrix on any $G$, we get a non-symmetric $R$-matrix on $G^2$. I guess this also works for any symmetric monoidal category $\mathcal{C}$, that $\mathcal{C}\otimes \mathcal{C}$ has a braided structure. $\endgroup$
    – Pulcinella
    Commented Aug 6, 2023 at 11:57

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