Radicals of co-quasitriangular map

Question

Let $B=\mathcal{O}_R\left(GL(n)\right)$ be a localization of the algebra $A(R)$ of functions on the quantum formal group corresponding to the matrix $R$ ["Quantization of Lie groups and Lie algebras", Faddev, Reshetikhin, Takhtajan] at $\mathrm{det}$ (quantum determinant), where $\mathrm{det}$ is a central grouplike element of degree $n$, which is not a zero divisor and $\mathrm{det}\ a=f(a)\mathrm{det}$ for some $f\in\mathrm{Aut}(A(R))$. We assume that $A(R)$ is defined with respect to $n\times n$ matrix coalgebra $C$ with standard coproduct $\Delta\left(t_{\ j}^{i}\right)=t_{\ k}^{i}\otimes t^{k}_{\ j}$, where $\left(t^{i}_{\ j}\right)$ forms basis of $C$ and there exist a bilinear $r:C\otimes C\rightarrow k$ s.th. $r\left(t^{i}_{\ j},t^{k}_{\ l}\right)=R^{i \ k}_{\ j \ l}$, where $R=\left(R^{i \ k}_{\ j \ l}\right)$ is a bi-invertible solution of Yang-Baxter equation on $k^n\otimes k^n$. One can define antipode map $S$ and co-quasitriangular structure $\textbf{r}$ on $B$.

In one article I found a statement that if $\mathrm{det}-1$ belongs to the left and right radicals of $\textbf{r}$ then we can define some additional structures.

My question is : What is a radical of map which gives us a co-quasitriangular structure on $B$ ? I can't find this definition.

Answer 1

The co-quasitriangular structure $r$ is a bilinear map over the ground field. Hence one can define right and left radicals as in linear algebra.

Note that the quantum determinant is a sum, e.g. in an example presented in the paper by Faddev-Rheshitikhin-Takhtajan, $ \operatorname{det}_q T= \sum_{s∈S_n} (-q)^{l(s)} t^1_{s_1}\ldots t^n_{s_n} $. For this reason one needs to extend the map $r$ to $A(R)\otimes A(R)$ first. This can be done using the axioms of a dual $r$-matrix. This way one obtains a dual $r$-matrix on $A(R)\otimes A(R)$ and the left and right radical are defined. The condition then becomes that $r(\operatorname{det}_q-1\otimes a)=0$ for any $a\in A(R)$. If this holds, i.e. $\operatorname{det}-1$ is in the radical, then the map $r$ descents to a dual $r$-matrix on the quotient $A(R)/(\operatorname{det}-1)$ which can be viewed as functions $\mathcal{O}_R(SL_n)$ on quantum $SL_n$.