Action of left $\mathbb{C}_q[SL_2]$-crossed modules - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T23:25:43Z http://mathoverflow.net/feeds/question/116080 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/116080/action-of-left-mathbbc-qsl-2-crossed-modules Action of left $\mathbb{C}_q[SL_2]$-crossed modules Arm Boris 2012-12-11T13:26:51Z 2013-05-13T17:22:00Z <p>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&amp;e_b\ e_c&amp;e_d\<br> \end{array} \right)= \left(\begin{array}{cc} q e_a +q\mu^2 e_d &amp;e_b\ e_c&amp;q^{-1}e_d\<br> \end{array} \right) \end{eqnarray}</p> <p>\begin{eqnarray} \nonumber b\triangleright \left(\begin{array}{cc} e_a&amp;e_b\ e_c&amp;e_d\<br> \end{array} \right)= \left(\begin{array}{cc} \mu e_c &amp;\mu e_d\ 0&amp;0\<br> \end{array} \right) \end{eqnarray}</p> <p>\begin{eqnarray} \nonumber c\triangleright \left(\begin{array}{cc} e_a&amp;e_b\ e_c&amp;e_d\<br> \end{array} \right)= \left(\begin{array}{cc} \mu e_b &amp;0\ q\mu e_d&amp;0\<br> \end{array} \right) \end{eqnarray} \begin{eqnarray} \nonumber d\triangleright \left(\begin{array}{cc} e_a&amp;e_b\ e_c&amp;e_d\<br> \end{array} \right)= \left(\begin{array}{cc} q^{-1} e_a &amp;e_b\ e_c&amp;q e_d\<br> \end{array} \right) \end{eqnarray}</p> <p>My question is how (or where) can we find the details of this result ? Thank you</p> http://mathoverflow.net/questions/116080/action-of-left-mathbbc-qsl-2-crossed-modules/118257#118257 Answer by Réamonn Ó Buachalla for Action of left $\mathbb{C}_q[SL_2]$-crossed modules Réamonn Ó Buachalla 2013-01-07T09:24:10Z 2013-01-07T09:24:10Z <p>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 ${\bf z}=q^2a+d-(q^3+q^{-1})$. </p> <p>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.</p>