Orthogonality of root spaces and nondegeneracy on the Cartan subalgebra

We first justify the zero-weight convention in (i). By [L3], $\mathfrak{h}$ 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 [L2]. For every $H\in \mathfrak{h}$, the equality $0=[H,x]={\sum }_{\gamma }\gamma (H)x_{\gamma }$ and directness of the decomposition give $\gamma (H)x_{\gamma }=0$ for every $\gamma$ and every $H$. Since each root $\gamma$ is a nonzero functional, this forces every $x_{\gamma }=0$, so $x=H_0\in \mathfrak{h}$ and ${\mathfrak{g}}_0=\mathfrak{h}$. Now let $x\in {\mathfrak{g}}_{\alpha }$ and $y\in {\mathfrak{g}}_{\beta }$. Invariance [L1] gives $0=B([H,x],y)+B(x,[H,y])=(\alpha (H)+\beta (H))B(x,y)$. If $\alpha +\beta \mathrm{\ne }0$, some $H\in \mathfrak{h}$ has $(\alpha +\beta )(H)\mathrm{\ne }0$, whence $B(x,y)=0$. This proves (i).

For (ii) let $x\in \mathfrak{h}$ satisfy $B(x,\mathfrak{h})=0$. By (i) every ${\mathfrak{g}}_{\gamma }$ with $\gamma \mathrm{\ne }0$ is orthogonal to $\mathfrak{h}={\mathfrak{g}}_0$, so $B(x,{\mathfrak{g}}_{\gamma })=0$ for all $\gamma \mathrm{\ne }0$ as well, and by [L2] $B(x,\mathfrak{g})=0$. Nondegeneracy [L4] gives $x=0$.