How to compute the inverse of a quantum determinant?

Question

Let $X=(x_{ij})_{mn}$ be a quantum matrix with the commutation relations between entries: \begin{alignat*}{2} & x_{ij} x_{il} = q x_{il} x_{ij}, && j < l, \\ & x_{ij} x_{kj} = q x_{kj} x_{ij}, && i<k, \\ & x_{ij} x_{kl} = x_{kl} x_{ij}, && i<k, j>l, \\ & x_{ij} x_{kl} = x_{kl} x_{ij} + (q - q^{-1})x_{il}x_{kj}, \quad && i < k, j < l. \end{alignat*} For $n \in \mathbb{Z}_{>0}$, denote $[n]=\{1, \ldots, n\}$. For any $I \subset [m]$, $J \subset [n]$, $|I|=|J|=l \in \mathbb{Z}_{>0}$, a quantum minor $\Delta_{I,J}$ is defined as follows \begin{align*} \Delta_{I,J} = \sum_{\sigma \in S_l} (-q)^{\ell(\sigma)} x_{i_1, j_{\sigma(1)}} \cdots x_{i_l, j_{\sigma(l)}}, \end{align*} where $\{i_1 < \cdots < i_l\} = I$, $\{j_1< \cdots < j_l\}=J$, and $\ell(\sigma)$ is the length of the permutation $\sigma$.

When $m=n$ and $I=[n]$, $\Delta_{I,I}=\det_q(X)$ is the quantum determinant of $X$. How to compute $(\det_q(X))^{-1}$. Are there some references about this? Thank you very much.

Answer 1

Algebraic properties of Manin matrices, page 8: $(\det_q M)^{-1}=\det_q(M^{-1})$.

Answer 2

The determinant $\text{det}_q(X)$ is not an invertible element of the algebra $X$. To see this look at the case of $X$, for $n=2$, and try to formulate an argument in terms of the standard basis $$ X_{11}^a X_{12}^b X_{21}^c X_{22}^d. $$ Think of the $q=1$ case, where the algebra $X$ reduces to the coordinate algebra of $M_2(\mathbb{C})$. The determinant function will vanish on all matrices with trivial determinant, and so, cannot admit an inverse/

For $\text{det}$ to be invertible, you need to restrict to GL$(\mathbb{C})$. Dually this corresponds to adjoining an inverse to $\text{det}$. In the $q$-deformed case we mimic this and construct the quantum general linear group by adjoining a inverse. Explicitly, since $det_q(X)$ is well-known to be a central element of $X$, it is easy to attach an inverse viewed as an Ore extension, or equivalently just add an extra generator $\text{det}_q$ along with the obvious additional relations.

The quantum special linear group can then be constructed from the quantum general linear group by quotienting by the ideal generated by $\text{det}_q(X) - 1$. Alternatively, one can go straight to the quantum special linear group by quotienting $X$ by the ideal generated by $\text{det}_q(X) - 1$.

This is of course closely related to the fact $X$ does not admit a Hopf algebra structure, but the quantum general linear, and special, groups do. This is all explained in Klimyk and Schmudgen's text, for example.