I was reading this paper, and I want to understand the definition of $Fix_{G_k}{\mathcal{W}}$ given in $1.2$.
I tried to use the formula on a specific example and it did not worked (I think that I did not understand the definition of Fix).
Let $G=S_3$ and the signature $\sigma=(1;3)$. This signature has two possible generating vectors $\mathcal{V}=(\alpha_1,\beta_1,g_1)$ (See paper) given by:
\begin{align} \alpha_1= (1 2), \beta_1=(13), \text{ } g_1=(123)\\ \alpha_1= (1 2), \beta_1=(123),g_1=(123) \end{align}
On both cases, $G_1$ is the conjugaccy class associated to the cyclic group generated by $g_1=(123)$
Let $\mathcal{W}$ be the irreducible two dimensional representation of $\rho :S_3 \to GL_2(\mathbb{C})$ given by: \begin{align} \rho_{(12)}=\begin{pmatrix} -1 & -1 \\ 0 & 1 \\ \end{pmatrix} , \rho_{(123)}=\begin{pmatrix} -1 & -1 \\ 1 & 0 \\ \end{pmatrix} \end{align}
I want to know if $Fix_{G_1}{\mathcal{W}}$ means the fixed space for the matrix $\rho_{(123)}$. Since that would mean that is the zero vector. It would follow that $\dim Fix_{G_1}{\mathcal{W}}=0$ and since $\gamma=1$;$\dim \mathcal{W}=2$ and $n_1=3$. Hence $\dim B=3$, but this is not possible since in the paper they proved that if $G=S_3$ and $\sigma=(1;3)$ then the dimension of the factors $B_i$ have dimension $\le 1$.
I tried to do the same for other examples and the same contradiction happens. This is why I think that I did not understand the definition of $Fix_{G_k} \mathcal{W}$ and is probably refering to something different.