Opposite root spaces bracket to the Killing-dual line

By [L0] and Brackets of root spaces add their roots, the bracket $[x,y]$ lies in ${\mathfrak{g}}_0=\mathfrak{h}$. For every $h\in \mathfrak{h}$, invariance of the Killing form gives $B([x,y],h)=B(x,[y,h])=\alpha (h)B(x,y)=B(x,y)B(H_{\alpha },h)$.

Because both $[x,y]$ and $B(x,y)H_{\alpha }$ lie in $\mathfrak{h}$ and have the same Killing pairings with every $h\in \mathfrak{h}$, nondegeneracy of the restriction from Under Choice, the Killing form pairs only opposite root spaces forces $[x,y]=B(x,y)H_{\alpha }$.

The root space ${\mathfrak{g}}_{\alpha }$ is nonzero by the definition of a root. Choose $0\mathrm{\ne }x\in {\mathfrak{g}}_{\alpha }$; the perfect opposite-root pairing in [L1] supplies $y\in {\mathfrak{g}}_{-\alpha }$ with $B(x,y)\mathrm{\ne }0$. Also $H_{\alpha }\mathrm{\ne }0$, for otherwise its defining equation would make the nonzero root $\alpha$ the zero functional. Thus step 2.1 gives a nonzero scalar multiple of $H_{\alpha }$ in the bracket image, while every bracket lies on that line. Hence $[{\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{-\alpha }]=\mathbb{C}{H}_{\alpha }$.