Question

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 tha 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}

My question is how (or where) can we find the details of this result ? Thank you

Answer 1

There's nothing too difficult going on here. The left module relations you describe for $\Lambda_{SL_N} =$span{$e_a,e_b,e_c,e_d$} can be seen to come from the construction of $\Lambda^1_{SU_N}$ as a quotient $C_q[SL_N]^+/I$, where $I$ is the ideal (from the famous Woronowicz paper) generated by the elements \begin{eqnarray*} b^2,~c^2,~b(a-d),~c(a-d), {\bf z} b, {\bf z} c,~{\bf z}(a-d), \end{eqnarray*} $$ ~ {\bf z}(q^2a+d-(q^2+1))\ ~a^2+q^{2}d^2-(1+q^2)(ad+q^{-1} bc), $$ for ${\bf z}=q^2a+d-(q^3+q^{-1})$.

The basis they use is the obvious one, and the relations follow from basic algebraic manipulation. (More generally, one can describe the ideal using the quantum Killing form for the standard coquasi-triangular structure of $C_q[SL_N]$, but this is not at all necessary for this example.) I'll try to add more detail later.