Quantum planes and quantum matrices.

Question

Let $A = \mathbb{C}_q\langle x,y\rangle/(xy-qyx)$ be a quantum plane. Let $M = \left( \begin{matrix} a & b \\ c & d \end{matrix} \right)$. If we require that $x'y' = qy'x'$, where $\left( \begin{matrix} x' \\ y' \end{matrix} \right) = M\left( \begin{matrix} x \\ y \end{matrix} \right)$. Then we have $$ (ax+by)(cx+dy) = x'y' = qy'x' = q(cx+dy)(ax+by). $$ It follows that $$ acx^2+bcyx+adxy+bdy^2 = qcax^2+qdayx+qcbxy+qdby^2. $$ Therefore $ac=qca, bd=qdb, bc+qad=qda+q^2cb$. But we cannot deduce that $bc=cb$. How to obtain $bc=cb$? Thank you very much.

Answer 1

This is e.g. explained in [Kassel: Quantum groups, Theorem IV.3.1]. One additionally uses $$\begin{pmatrix}x''\\y''\end{pmatrix}=\begin{pmatrix}a&c\\b&d\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$$ and $y''x''=qx''y''$.

This first yields $ad-da=q^{-1}cb-qbc$. Together with your last equation divided by $q$ this gives $(q+q^{-1})(bc-cb)=0$. Plugging in the usual assumption that $q^2\neq -1$, one gets the claimed relation.