Roots of a complex semisimple Lie algebra form a reduced crystallographic root system

Properties (i) and (v) are [L1] and [L4]; property (iii) is [L2] together with [L5], which also shows $-\alpha \in \mathrm{\Phi }$; and (iv) is [L3].

For (ii): if $H\in \mathfrak{h}$ has $\alpha (H)=0$ for every $\alpha \in \mathrm{\Phi }$, then $[H,{\mathfrak{g}}_{\alpha }]=\alpha (H){\mathfrak{g}}_{\alpha }=0$ for every root and $[H,\mathfrak{h}]=0$ because $\mathfrak{h}$ is abelian; by the direct sum of [L1] this gives $[H,\mathfrak{g}]=0$, so $H$ is central and $H=0$ by [L6]. Hence the common kernel is zero, and therefore $\mathrm{\Phi }$ spans ${\mathfrak{h}}^{\ast }$: a finite set of functionals spans the dual space exactly when no nonzero vector is annihilated by all of them, applied to the dual pairing between $\mathfrak{h}$ and ${\mathfrak{h}}^{\ast }$.

Let ${\mathfrak{h}}_{\mathbb{R}}={\mathrm{span}}_{\mathbb{R}}\{h_{\alpha }:\alpha \in \mathrm{\Phi }\}$. The coroots span $\mathfrak{h}$ over $\mathbb{C}$: by step 1.2 the roots span ${\mathfrak{h}}^{\ast }$, their Killing-duals therefore span $\mathfrak{h}$, and [L7] says that each $H_{\alpha }$ is a nonzero complex multiple of $h_{\alpha }$. For $H\in {\mathfrak{h}}_{\mathbb{R}}$ every root value $\beta (H)$ is real, because it is a real linear combination of the integers $\beta (h_{\alpha })$ from [L4]. If also $H\in i{\mathfrak{h}}_{\mathbb{R}}$, every $\beta (H)$ is both real and purely imaginary, hence zero; step 1.2 gives $H=0$. The complex spanning and this zero intersection prove $\mathfrak{h}={\mathfrak{h}}_{\mathbb{R}}\oplus i{\mathfrak{h}}_{\mathbb{R}}$ as real vector spaces.

For $H,K\in {\mathfrak{h}}_{\mathbb{R}}$, the root-space decomposition and one-dimensionality in [L1] give $B(H,K)=\mathrm{tr}(\mathrm{ad}H\mathrm{ad}K)={\sum }_{\beta \in \mathrm{\Phi }}\beta (H)\beta (K)\in \mathbb{R}$. Thus $B(H,H)={\sum }_{\beta \in \mathrm{\Phi }}\beta (H{)}^2\ge 0$, and equality forces every $\beta (H)=0$, hence $H=0$ by step 1.2. Therefore $B{\mathrm{\mid }}_{{\mathfrak{h}}_{\mathbb{R}}}$ is positive definite. In particular $B(h_{\alpha },h_{\alpha })=4\mathrm{/}\alpha (H_{\alpha })>0$, so $H_{\alpha }$ is a positive real multiple of $h_{\alpha }$. It follows that the Killing-dual map sends $E={\mathrm{span}}_{\mathbb{R}}\mathrm{\Phi }$ isomorphically onto ${\mathfrak{h}}_{\mathbb{R}}$. Transporting $B$ across that map defines a positive-definite inner product on $E$.

For the inner product of step 3.1, $2(\beta ,\alpha )\mathrm{/}(\alpha ,\alpha )=2B(H_{\beta },H_{\alpha })\mathrm{/}B(H_{\alpha },H_{\alpha })=\beta (h_{\alpha })$. Hence the map $\beta \mapsto \beta -\beta (h_{\alpha })\alpha$ of [L3] is precisely the orthogonal reflection in ${\alpha }^{\perp }$, and the Euclidean coroot maps to $h_{\alpha }$. Together with finiteness and spanning by the definition of $E$, [L2] gives reducedness, [L3] reflection stability, and [L4] crystallographic integrality. Thus $\mathrm{\Phi }\subset E$ satisfies every reduced crystallographic root-system axiom, not merely properties (i)–(v). The Axiom of Choice is inherited through [L1], [L6], and [L7].