sl2 subalgebra associated with a root (${\mathfrak{m}}_{\alpha }\cong {\mathfrak{sl}}_2$)

For every root $\alpha$ of a complex semisimple Lie algebra, there are root vectors $e_{\alpha }\in {\mathfrak{g}}_{\alpha }$ and $f_{\alpha }\in {\mathfrak{g}}_{-\alpha }$ and an element $h_{\alpha }\in \mathfrak{t}$ satisfyingTheir span is an ${\mathfrak{sl}}_2$ subalgebra.

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 $\alpha (t_{\alpha })\ne 0$. 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 $\mathrm{ad}{t}_{\alpha }$ nilpotent, while the root-space decomposition makes it diagonalizable, forcing $t_{\alpha }$ into the zero center of a Lie algebra.