The bracket of opposite root spaces is the root line

We first prove that the zero-weight space in [L1] is $\mathfrak{h}$. By [L4], $\mathfrak{h}$ is abelian, so $\mathfrak{h}\subseteq {\mathfrak{g}}_0$. Conversely, if $x\in {\mathfrak{g}}_0$, write $x=H_0+{\sum }_{\gamma \in \mathrm{\Phi }}{x}_{\gamma }$ by [L1]. For every $H\in \mathfrak{h}$, directness and $0=[H,x]={\sum }_{\gamma }\gamma (H)x_{\gamma }$ imply $\gamma (H)x_{\gamma }=0$ for every root $\gamma$. Since each $\gamma$ is a nonzero functional, every $x_{\gamma }$ vanishes; hence $x=H_0\in \mathfrak{h}$ and ${\mathfrak{g}}_0=\mathfrak{h}$. In particular [L1] gives $[{\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{-\alpha }]\subseteq \mathfrak{h}$.

Let $e\in {\mathfrak{g}}_{\alpha }$, $f\in {\mathfrak{g}}_{-\alpha }$ and $H\in \mathfrak{h}$. By step 1.1, $[e,f]\in \mathfrak{h}$, and invariance [L2] together with $[f,H]=\alpha (H)f$ gives $B([e,f],H)=B(e,[f,H])=\alpha (H)B(e,f)=B(B(e,f)H_{\alpha },H)$. Nondegeneracy of $B{\mathrm{\mid }}_{\mathfrak{h}}$ yields $[e,f]=B(e,f)H_{\alpha }$, so every such bracket lies in $\mathbb{C}{H}_{\alpha }$. By [L3] some $e,f$ have $B(e,f)\mathrm{\ne }0$; then their bracket is nonzero because $H_{\alpha }\mathrm{\ne }0$ by Killing-dual vector of a root. Therefore the bracket is exactly $\mathbb{C}{H}_{\alpha }$.