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For an arbitrary finite-dimensional complex Lie algebra $L$, a Cartan subalgebra is a nilpotent Lie subalgebra $H$ equal to its own normalizer. When $L$ is semisimple, this is equivalently a maximal abelian subalgebra consisting of elements that act semisimply in the Adjoint representation.

Choose such an $H$. Simultaneous diagonalization gives the root-space decompositionThe nonzero weights $\alpha \in H^{\ast }$ are the roots. The restriction of the Killing form $B$ to $H$ is a nondegenerate bilinear form, so each $\alpha$ corresponds to a unique $t_{\alpha }\in H$ with $\alpha (h)=B(t_{\alpha },h)$. On the real span of these $t_{\alpha }$, the restriction of $B$ supplies a positive-definite inner product after choosing the standard real form. The sl2 subalgebra associated with a root givesand shows that these reflections preserve the finite set $\Phi$. Thus the roots form a finite reduced crystallographic root system, whose Weyl group is generated by these reflections.

A torus $\mathfrak{t}$ in a Lie algebra is an abelian subalgebra whose elements act semisimply in the Adjoint representation. Its weight-space decomposition is

Invariance of the nondegenerate Trace form of a Lie algebra representation givesNondegeneracy 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 $h_{\alpha }\in \mathfrak{t}$ satisfyingTheir span is a copy of ${\mathfrak{sl}}_2$.

SetThen $\dim \mathfrak{h}=\dim \mathfrak{t}-1$, $\mathfrak{t}=\mathbb{C}{h}_{\alpha }\oplus \mathfrak{h}$, and every element of $\mathfrak{h}$ commutes with $e$, $f$, and $\mathfrak{t}$. Hence $\mathfrak{h}$ is abelian andas a direct sum of commuting Lie algebras. This is the two-root decomposition with a nondegenerate trace form.

Use the nondegenerate restriction of the Killing form to $\mathfrak{t}$ to define $t_{\alpha }\in \mathfrak{t}$ byKilling-form invariance and the root-space decomposition show that ${\mathfrak{g}}_{\alpha }$ pairs nondegenerately with ${\mathfrak{g}}_{-\alpha }$ and orthogonally with every other root space. Choose nonzero $e\in {\mathfrak{g}}_{\alpha }$ and $f\in {\mathfrak{g}}_{-\alpha }$ with $\kappa (e,f)\ne 0$. For $t\in \mathfrak{t}$,so

We need $\alpha (t_{\alpha })\ne 0$. If it were zero, the span of $e,f,t_{\alpha }$ would be a solvable Heisenberg-type Lie algebra with central commutator $[e,f]$. By Lie theorem, its adjoint action on $\mathfrak{g}$ can be upper triangularized, so $\mathrm{ad}[e,f]$ is nilpotent. But $[e,f]\in \mathfrak{t}$, and elements of the Cartan subalgebra act semisimply; hence $\mathrm{ad}[e,f]=0$. A semisimple Lie algebra has zero center, contradicting $[e,f]\ne 0$.

SetRescale $f$ so that $[e_{\alpha },f_{\alpha }]=h_{\alpha }$. Since $e_{\alpha }$ and $f_{\alpha }$ lie in the $\alpha$ and $-\alpha$ root spaces,Thus ${\mathfrak{m}}_{\alpha }=⟨e_{\alpha },h_{\alpha },f_{\alpha }⟩$ is the sl2 subalgebra associated with a root.

For a complex semisimple Lie algebra, restrict the module $\mathfrak{h}\oplus {⨁}_{j\ne 0}{\mathfrak{g}}_{j\alpha }$ to the sl2 subalgebra associated with a root. All its weights are even. Its raising operator maps the weight-zero space $\mathfrak{h}$ onto the one-dimensional line of the chosen root vector, so there is exactly one nontrivial irreducible summand: the adjoint module of highest weight $2$. Thus the root spaces at $\pm \alpha$ are one-dimensional and $2\alpha$ is not a root. For any parallel root $c\alpha$, the integers $2c$ and $2/c$ have product $4$; excluding double and half roots leaves only $c=\pm 1$.

For nonproportional roots $\gamma ,\delta$ of a reduced crystallographic root system, $S_{\gamma ,\delta }=\{\delta -p\gamma ,\dots ,\delta +q\gamma \}$ is consecutive, and $p-q=⟨\delta ,{\gamma }^{\vee }⟩$. Realize the crystallographic root system as the roots of a complex semisimple Lie algebra. The sum of the one-dimensional root spaces in this string is a module for the sl2 subalgebra associated with a root. Its weights are $⟨\delta ,{\gamma }^{\vee }⟩+2n$; they have one fixed parity. Complete reducibility gives a sum of irreducible representations of the sl2 Lie algebra. Two summands of this parity would overlap at weight zero when the parity is even, or at weights $\pm 1$ when it is odd, contradicting the one-dimensional root spaces. Thus there is one irreducible string, whose opposite endpoint weights sum to zero. This proves the relation. For distinct simple roots, $p=0$ and the cardinality is $1-⟨\delta ,{\gamma }^{\vee }⟩$. The nonproportional hypothesis excludes $\delta =\pm \gamma$, where zero is not a root and the consecutive-string statement fails.

For the symplectic Lie algebra $\mathfrak{sp}(4)$ preserving $J=\left(\begin{array}{cc}0& I_2\\ -I_2& 0\end{array}\right)$, write $h_1=E_{11}-E_{33}$ and $h_2=E_{22}-E_{44}$. Its positive roots of a root system have sl2 subalgebra associated with a root triplesThe matrix unit commutator identity verifies $[X_{\alpha },Y_{\alpha }]=H_{\alpha }$ and the two eigenvalue relations $[H_{\alpha },X_{\alpha }]=2X_{\alpha }$, $[H_{\alpha },Y_{\alpha }]=-2Y_{\alpha }$.