A torus $\mathfrak t$ in a Lie algebra is an abelian subalgebra whose elements act semisimply in the <Adjoint representation>. Its <root-space decomposition>[weight-space decomposition] is
$$
\mathfrak g=\bigoplus_{\gamma\in\mathfrak t^*}\mathfrak g_\gamma,\qquad
\mathfrak g_\gamma=\{x:[t,x]=\gamma(t)x\text{ for every }t\in\mathfrak t\}.
$$

Invariance of the nondegenerate <Trace form of a Lie algebra representation> gives
$$
(\mathfrak g_\gamma,\mathfrak g_\delta)_V=0
\quad\text{unless}\quad \gamma+\delta=0.
$$
Nondegeneracy therefore forces $\beta=-\alpha$. The standard <sl2 subalgebra associated with a root> argument gives one-dimensional opposite root spaces with vectors $e\in\mathfrak g_\alpha$, $f\in\mathfrak g_{-\alpha}$ and $h_\alpha\in\mathfrak t$ satisfying
$$
[h_\alpha,e]=2e,\qquad [h_\alpha,f]=-2f,\qquad[e,f]=h_\alpha.
$$
Their span is a copy of $\mathfrak{sl}_2$.

Set
$$
\mathfrak h=\ker(\alpha:\mathfrak t\to\mathbb C).
$$
Then $\dim\mathfrak h=\dim\mathfrak t-1$, $\mathfrak t=\mathbb Ch_\alpha\oplus\mathfrak h$, and every element of $\mathfrak h$ commutes with $e$, $f$, and $\mathfrak t$. Hence $\mathfrak h$ is abelian and
$$
\mathfrak g=\mathfrak{sl}_2\oplus\mathfrak h
$$
as a direct sum of commuting Lie algebras. This is the <two-root decomposition with a nondegenerate trace form>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 102 / 2 / ii / Solution

Let a maximal torus $\mathfrak t$ give a decomposition $\mathfrak g=\mathfrak g_\alpha\oplus\mathfrak t\oplus\mathfrak g_\beta$. If $\mathfrak g$ has a nondegenerate trace form, then $\beta=-\alpha$, the opposite root spaces generate $\mathfrak{sl}_2$, and
$$
\mathfrak g\cong\mathfrak{sl}_2\oplus\ker\alpha,
$$
where $\ker\alpha\subset\mathfrak t$ is central and abelian.


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