Finding expression for $([[e_r^{\ast},e_k], e_s^{\ast}], e_l).$ 
Let $(\mathfrak {g}, \mathfrak {g}_+, \mathfrak {g}_-)$ be a finite dimensional Manin triple i.e. $\mathfrak g$ is a finite dimensional Lie algebra endowed with a non-degenerate invariant bilinear form $\left \langle \cdot, \cdot \right \rangle$ and $\mathfrak g_+$ and $\mathfrak g_-$ are Lie subalgebras such that $\mathfrak g = \mathfrak g_+ \oplus \mathfrak g_-$ as vector spaces with both of $\mathfrak {g}_+$ and $\mathfrak {g}_-$ being isotropic with respect to the bilinear form $\left \langle \cdot, \cdot \right \rangle$ i.e. the bilinear form is trivial when restricted to $\mathfrak {g}_+$ and $\mathfrak {g}_-.$ In fact, in this case $\mathfrak g_+$ and $\mathfrak g_-$ are maximal isotropic with respect the bilinear form $\left \langle \cdot, \cdot \right \rangle$ (such subalgebras are called Lagrangian subalgebras). Hence the non-degenerate bilinear form $\left \langle \cdot, \cdot \right \rangle$ induces a non-degenerate pairing $(\cdot, \cdot) : \mathfrak g_- \otimes \mathfrak g_+ \longrightarrow \mathbb R$ and consequently it gives rise to a vector space isomorphism $\mathfrak g_+^{\ast} \simeq \mathfrak g_{-}.$ Thus $\dim \mathfrak {g}$ is even and $\dim \mathfrak {g}_+ = \dim \mathfrak {g}_- = \frac {1} {2} \dim \mathfrak {g}.$ Now let $(e_i)_{1}^{n}$ be a basis of $\mathfrak g_+$ and $(e_i^{\ast})_{1}^{n}$ be the dual basis of $\mathfrak g_+^{\ast} \simeq \mathfrak g_-.$ Let $[e_i,e_j] = \sum\limits_{s} \alpha_{ij}^s e_s$ and $[e_i^{\ast}, e_j^{\ast}] = \sum\limits_{s} \beta_s^{ij} e_s^{\ast}.$ Then how to compute $([[e_r^{\ast}, e_k], e_s^{\ast}], e_l)$ in terms of $\alpha_{ij}^s$ and $\beta_s^{ij}\ $? In Etingof's lecture notes on quantum groups it is claimed that $$([[e_r^{\ast}, e_k], e_s^{\ast}], e_l) = \sum\limits_{t} \alpha_{ kt}^{r} \beta_l^{ts} + \sum\limits_{t} \alpha_{tl}^{s} \beta_k^{rt}.$$

But I am having hard time obtaining this equality. First of all what is $[e_r^{\ast}, e_k]\ $? Could anyone please help me in this regard?
Thanks for your time.
Source $:$ Lecture Notes on Quantum Groups by Pavel Etingof and Oliver Schiffmann (Lecture $4,$ Page Nos. $34 - 35$).


 A: Let me compute $[e_r^{\ast}, e_k].$ First write $[e_r^{\ast}, e_k] = [e_r^{\ast}, e_k]_+ + [e_r^{\ast}, e_k]_-$ for $[e_r^{\ast}, e_k]_+ \in \mathfrak {g}_+$ and $[e_r^{\ast}, e_k]_- \in \mathfrak {g}_-.$ Upon identifying $\mathfrak {g}_-$ with $\mathfrak {g}_+^{\ast}$ via the pairing $(\cdot, \cdot)$ we find that $[e_r^{\ast}, e_k]_+$ is determined by the action of the basis elements of $\mathfrak {g}_+^{\ast} \simeq \mathfrak {g}_-$ on it. So we have $$\begin{align*} (e_s^{\ast}, [e_r^{\ast}, e_k]_+) & = (e_s^{\ast}, [e_r^{\ast}, e_k])\ \ (\because \mathfrak {g}_-\ \text {is isotropic}) \\ & =([e_s^{\ast}, e_r^{\ast}], e_k)\ \ (\because (\cdot, \cdot)\ \text {is invariant}) \\ & = \left (\sum\limits_{t} \beta^{sr}_{t} e_t^{\ast}, e_k \right ) \\ & = \beta_{k}^{sr} \end{align*}$$ This shows that $[e_r^{\ast}, e_k]_+ = \sum\limits_{s} \beta_k^{sr} e_s.$ Similarly $[e_r^{\ast}, e_k]_-$ can be found by it's action on the basis elements of $\mathfrak {g}_+$ and we have $$\begin{align*} ([e_r^{\ast}, e_k]_-, e_s) & = ([e_r^{\ast}, e_k], e_s)\ \ (\because \mathfrak {g}_+\ \text {is isotropic}) \\ & = (e_r^{\ast}, [e_k, e_s])\ \ (\because (\cdot, \cdot)\ \text {is invariant}) \\ & = \left (e_r^{\ast}, \sum\limits_{t} \alpha_{ks}^{t} e_{t} \right ) \\ & = \alpha_{ks}^{r} \end{align*}$$
This shows that $[e_r^{\ast}, e_k]_{-} = \sum\limits_{s} \alpha_{ks}^{r} e_s^{\ast}.$ Thus we have $$[e_r^{\ast}, e_k] = \sum\limits_{s} \beta_{k}^{sr} e_{s} + \sum\limits_{s} \alpha_{ks}^{r} e_{s}^{\ast}.$$
