= Nonisotropic root lemma
{title2=$\alpha(t_\alpha)\ne0$}
For a <root of a root system> of a complex <semisimple Lie algebra>, define $t_\alpha$ by $B(t_\alpha,h)=\alpha(h)$ using the <Killing form> on its <Cartan subalgebra>. Then $\alpha(t_\alpha)\ne0$. Otherwise vectors $x\in\mathfrak g_\alpha$, $y\in\mathfrak g_{-\alpha}$ with $B(x,y)=1$ would generate a <Solvable Lie algebra> with $[x,y]=t_\alpha$ central. The <Lie theorem> would make $\operatorname{ad}t_\alpha$ nilpotent, while the <root-space decomposition> makes it diagonalizable, forcing $t_\alpha$ into the zero <center of a Lie algebra>.
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