# Action of left $\mathbb{C}_q[SL_2]$-crossed modules

Shahn Majin and Xavier Gomez say in the beginig of their article (Noncommutative cohomology and electromagnetism on $\mathbb{C}_q [SL_2]$ at roots of unity) that the action of left $\mathbb{C}_q [SL_2]$-crossed modules is given by: \begin{eqnarray} \nonumber a\triangleright \left(\begin{array}{cc} e_a&e_b\\ e_c&e_d\\ \end{array} \right)= \left(\begin{array}{cc} q e_a +q\mu^2 e_d &e_b\\ e_c&q^{-1}e_d\\ \end{array} \right) \end{eqnarray}

\begin{eqnarray} \nonumber b\triangleright \left(\begin{array}{cc} e_a&e_b\\ e_c&e_d\\ \end{array} \right)= \left(\begin{array}{cc} \mu e_c &\mu e_d\\ 0&0\\ \end{array} \right) \end{eqnarray}

\begin{eqnarray} \nonumber c\triangleright \left(\begin{array}{cc} e_a&e_b\\ e_c&e_d\\ \end{array} \right)= \left(\begin{array}{cc} \mu e_b &0\\ q\mu e_d&0\\ \end{array} \right) \end{eqnarray} \begin{eqnarray} \nonumber d\triangleright \left(\begin{array}{cc} e_a&e_b\\ e_c&e_d\\ \end{array} \right)= \left(\begin{array}{cc} q^{-1} e_a &e_b\\ e_c&q e_d\\ \end{array} \right) \end{eqnarray} The quantum group $C_q[SL2]$ has a matrix of generators $$t^i_j= \begin{eqnarray} \nonumber \left(\begin{array}{cc} a&b\\ c&d\\ \end{array} \right) \end{eqnarray}$$ with relations $$ba = qab, ca = qac, db = qbd, dc = qcd, cb = bc, da − ad = qµbc, ad − q −1 bc = 1,$$ where $$µ = 1 − q^{-2}$$ , and the matrix coalgebra structure. My question is how (or where) can we find the details of this result ? Thank you

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