I am reading the book.
Let $G$ be a Poisson-Lie group and $r = \sum_{s,t} r^{st} X_s \otimes X_t \in g \wedge g$, where $g$ is the Lie algebra of $G$. In the end of page 60, the bracket on $\mathbb{C}[G]$ is given by \begin{align} \{f_1, f_2\} = \sum_{s,t} r^{st}( (X_s^L f_1)(X_t^L f_2) - (X_s^R f_1) (X_t^R f_2) ), \quad (1) \end{align} where $f_1, f_2 \in \mathbb{C}[G]$, $\{X_s\}$ is a basis of $g$ and $\{X_s^L\}$, (resp. $\{X_s^R\}$) are corresponding left (resp. right) invariant vector field on $G$.
In the case of $G = GL_n$, how to derive the formula $\{t_{ij}, t_{kl}\} = \sum_{a,b} (r^{ajbl} t_{ia}t_{kb} - r^{iakb} t_{aj}t_{bl})$ from (1)? This formula is on page 61 of the book of Chari and Pressley.
We have $X^L(t_{ij}) = (TX)_{ij}$, $X^R(t_{ij}) = (XT)_{ij}$.
Thank you very much.
Take a basis $\{E_{ab}: a,b=1,\ldots,n\}$ of $gl_n$, where $E_{ab}$ is a matrix whose $(a,b)$-entry is $1$ and $0$ elsewhere. Then we have $$ E_{ab}^L(t_{ij}) = (TE_{ab})_{ij} = \delta_{bj}t_{ia}, \\ E_{ab}^R(t_{ij}) = (E_{ab}T)_{ij} = \delta_{ia}t_{bj}. $$ Therefore $$ \{t_{ij}, t_{kl}\} \\ = \sum_{a,b,c,d} (r^{ab,cd} E_{ab}^L(t_{ij})E_{cd}^L(t_{kl}) - r^{ab,cd} E_{ab}^R(t_{ij})E_{cd}^R(t_{kl})) \\ = \sum_{a,b,c,d} (r^{ab,cd} \delta_{bj}t_{ia}\delta_{dl}t_{kc} - r^{ab,cd} \delta_{ia}t_{bj}\delta_{kc}t_{dl} \\ = \sum_{a,c} r^{ajcl} t_{ia}t_{kc} - \sum_{b,d} r^{ibkd} t_{bj}t_{dl} \\ = \sum_{a,b} (r^{ajbl} t_{ia}t_{kb} - r^{iakb} t_{aj}t_{bl}). $$