How to show that the structure constant on $\mathcal{G}^*$ is $C_{c}^{ab} = f_{cd}^b r^{ad} + f_{cd}^a r^{db}$? Let $(\mathcal{G}, \mathcal{G}^*, \delta)$ be a Lie bialgebra. Suppose that the structure constant on $\mathcal{G}^*$ and $\mathcal{G}$ are
\begin{align}
& [t^a, t^b]_* = C_c^{ab} t_c, \\
& [t_a, t_b] = f_{ab}^c t_c,
\end{align}
respectively.
Let $r = r^{ab} t_a \otimes t_b \in \mathcal{G} \otimes \mathcal{G}$ be a classical r-matrix and assume that $\delta: \mathcal{G} \to \mathcal{G} \wedge \mathcal{G}$ is given by $\delta(X) = [X \otimes 1 + 1 \otimes X, r]$, $X \in \mathcal{G}$. 
In the paper, it is said that the structure constant on $\mathcal{G}^*$ is $C_{c}^{ab} = f_{cd}^b r^{ad} + f_{cd}^a r^{db}$.
How to show that the structure constant on $\mathcal{G}^*$ is $C_{c}^{ab} = f_{cd}^b r^{ad} + f_{cd}^a r^{db}$?
Thank you very much.
 A: It is important to note that the literature often uses the shorthand $$[X\otimes 1+1\otimes X, Y\otimes Z]=(ad_X\otimes 1+1\otimes ad_X)(Y\otimes Z)=[X,Y]\otimes Z+X\otimes [Y,Z]$$ where $1$ is the identity map. We compute $C^{ab}_c=\langle [t^a,t^b]_*,t_c\rangle$ as follows. Recall that there is implied summation over repeated indices.
$\begin{align}\langle [t^a, t^b]_*,t_c\rangle &= \langle t^a \otimes  t^b ,[t_c\otimes 1 + 1 \otimes t_c,r]\rangle\\
 &=\langle t^a \otimes  t^b , [t_c\otimes 1 + 1 \otimes t_c,r^{de}t_d\otimes t_e ]\rangle\\
&=\langle t^a \otimes  t^b , r^{de}([t_c,t_d]\otimes t_e + t_d\otimes [t_c,\otimes t_e ])\rangle\\
&=\langle t^a \otimes  t^b , r^{de}(f_{cd}^f t_f\otimes t_e + f_{ce}^gt_d\otimes t_g)\rangle\\
&= r^{de}f_{cd}^f\langle t^a \otimes  t^b , t_f\otimes t_e\rangle + r^{de}f_{ce}^g\langle t^a \otimes  t^b ,t_d\otimes t_g\rangle\\
&= r^{de}f_{cd}^f\delta^a_f\delta^b_e + r^{de}f_{ce}^g\delta^a_d\delta^b_g\\
&= r^{db}f_{cd}^a+ r^{ae}f_{ce}^b\\
&= r^{db}f_{cd}^a+ r^{ad}f_{cd}^b\\
\end{align}
$
