[proofplan] The proof uses the root-space decomposition of $\mathfrak{g}$ and the orthogonality rule for the Killing form: $\mathfrak{g}_\alpha$ pairs nontrivially only with $\mathfrak{g}_{-\alpha}$. Since the Killing form is nondegenerate on the whole semisimple Lie algebra $\mathfrak{g}$, this forces the restricted pairing between opposite root spaces to be nondegenerate. To compute $[x,y]$, we first note that $[x,y] \in \mathfrak{h}$, then pair it against an arbitrary $h \in \mathfrak{h}$ and use [invariance of the Killing form](/theorems/3808) to identify it with $\kappa(x,y)t_\alpha$. [/proofplan]

[step:Use root-space orthogonality to isolate the only possible nonzero pairing]Fix $\alpha \in \Phi$. We use the standard root-space decomposition \begin{align*} \mathfrak{g} = \mathfrak{h} \oplus \bigoplus_{\beta \in \Phi} \mathfrak{g}_\beta \end{align*} and the orthogonality property of the Killing form for root spaces: if $\lambda,\mu \in \Phi \cup \{0\}$ and $\lambda + \mu \neq 0$, then the corresponding weight spaces are orthogonal with respect to $\kappa$ (citing a result not yet in the wiki: Killing-form orthogonality of root spaces). Here the $0$-weight space is $\mathfrak{h}$. Thus, for $x \in \mathfrak{g}_\alpha$, we have \begin{align*} \kappa(x,h) &= 0 &&\text{for every } h \in \mathfrak{h}, \\ \kappa(x,z) &= 0 &&\text{for every } z \in \mathfrak{g}_\beta \text{ with } \beta \neq -\alpha. \end{align*} Therefore $\mathfrak{g}_{-\alpha}$ is the only summand of the root-space decomposition on which an element of $\mathfrak{g}_\alpha$ can pair nontrivially.[/step]

[guided]Fix $\alpha \in \Phi$. The key structural input is that the Killing form respects the root-space decomposition in a very rigid way. We write \begin{align*} \mathfrak{g} = \mathfrak{h} \oplus \bigoplus_{\beta \in \Phi} \mathfrak{g}_\beta. \end{align*} The orthogonality theorem for root spaces says that if two weights $\lambda,\mu \in \Phi \cup \{0\}$ do not add to zero, then the corresponding weight spaces are orthogonal for the Killing form (citing a result not yet in the wiki: Killing-form orthogonality of root spaces). In this convention, the $0$-weight space is $\mathfrak{h}$. Now take $x \in \mathfrak{g}_\alpha$. Since $\alpha + 0 \neq 0$, the orthogonality theorem gives \begin{align*} \kappa(x,h) = 0 \end{align*} for every $h \in \mathfrak{h}$. Similarly, if $\beta \in \Phi$ and $\beta \neq -\alpha$, then $\alpha + \beta \neq 0$, so \begin{align*} \kappa(x,z) = 0 \end{align*} for every $z \in \mathfrak{g}_\beta$. Thus every possible nonzero pairing involving $x \in \mathfrak{g}_\alpha$ must occur against the single opposite root space $\mathfrak{g}_{-\alpha}$.[/guided]

[step:Deduce nondegeneracy of the opposite-root pairing from nondegeneracy on $\mathfrak{g}$]Let $x \in \mathfrak{g}_\alpha$ satisfy \begin{align*} \kappa(x,y) = 0 \end{align*} for every $y \in \mathfrak{g}_{-\alpha}$. By the previous step, $x$ is also orthogonal to $\mathfrak{h}$ and to every $\mathfrak{g}_\beta$ with $\beta \neq -\alpha$. Hence $x$ is orthogonal to every summand in \begin{align*} \mathfrak{g} = \mathfrak{h} \oplus \bigoplus_{\beta \in \Phi} \mathfrak{g}_\beta, \end{align*} so $\kappa(x,z)=0$ for every $z \in \mathfrak{g}$. Since $\mathfrak{g}$ is semisimple, Cartan's criterion gives that the Killing form $\kappa$ is nondegenerate on $\mathfrak{g}$ (citing a result not yet in the wiki: nondegeneracy of the Killing form on semisimple Lie algebras). Therefore $x=0$. The same argument with $\alpha$ replaced by $-\alpha$ shows that if $y \in \mathfrak{g}_{-\alpha}$ satisfies $\kappa(x,y)=0$ for every $x \in \mathfrak{g}_\alpha$, then $y=0$. Hence the bilinear pairing \begin{align*} \mathfrak{g}_\alpha \times \mathfrak{g}_{-\alpha} &\to k \\ (x,y) &\mapsto \kappa(x,y) \end{align*} is nondegenerate.[/step]

[guided]We now prove nondegeneracy in the first variable. Suppose $x \in \mathfrak{g}_\alpha$ satisfies \begin{align*} \kappa(x,y) = 0 \end{align*} for every $y \in \mathfrak{g}_{-\alpha}$. The previous step already showed that $x$ is orthogonal to every other summand in the root-space decomposition: it is orthogonal to $\mathfrak{h}$ and to each $\mathfrak{g}_\beta$ with $\beta \neq -\alpha$. Combining these facts, $x$ is orthogonal to every element of \begin{align*} \mathfrak{g} = \mathfrak{h} \oplus \bigoplus_{\beta \in \Phi} \mathfrak{g}_\beta. \end{align*} Equivalently, \begin{align*} \kappa(x,z)=0 \end{align*} for every $z \in \mathfrak{g}$. This is exactly where semisimplicity is used. For a finite-dimensional semisimple Lie algebra over a field of characteristic $0$, Cartan's criterion implies that the Killing form is nondegenerate on $\mathfrak{g}$ (citing a result not yet in the wiki: nondegeneracy of the Killing form on semisimple Lie algebras). Therefore the only vector orthogonal to all of $\mathfrak{g}$ is $0$, and hence $x=0$. The second variable is identical after replacing $\alpha$ by $-\alpha$. If $y \in \mathfrak{g}_{-\alpha}$ pairs to zero with every $x \in \mathfrak{g}_\alpha$, then the same argument shows $y=0$. Thus the restriction of $\kappa$ to \begin{align*} \mathfrak{g}_\alpha \times \mathfrak{g}_{-\alpha} \end{align*} is nondegenerate in both variables.[/guided]

[step:Show that the bracket of opposite root vectors lies in $\mathfrak{h}$] Let $x \in \mathfrak{g}_\alpha$ and $y \in \mathfrak{g}_{-\alpha}$. For every $h \in \mathfrak{h}$, the Jacobi identity gives \begin{align*} [h,[x,y]] &= [[h,x],y] + [x,[h,y]] \\ &= [\alpha(h)x,y] + [x,-\alpha(h)y] \\ &= \alpha(h)[x,y] - \alpha(h)[x,y] \\ &= 0. \end{align*} Thus $[x,y]$ belongs to the $0$-weight space for the adjoint action of $\mathfrak{h}$, which is $\mathfrak{h}$. Hence \begin{align*} [x,y] \in \mathfrak{h}. \end{align*} [/step]

[step:Identify the bracket by pairing against every element of $\mathfrak{h}$]Let $h \in \mathfrak{h}$. Using invariance of the Killing form, we compute \begin{align*} \kappa([x,y],h) &= \kappa(x,[y,h]). \end{align*} Since $y \in \mathfrak{g}_{-\alpha}$, we have \begin{align*} [h,y] = -\alpha(h)y, \end{align*} and therefore \begin{align*} [y,h] = \alpha(h)y. \end{align*} Substituting this into the previous identity yields \begin{align*} \kappa([x,y],h) &= \kappa(x,\alpha(h)y) \\ &= \alpha(h)\kappa(x,y). \end{align*} By the defining property of $t_\alpha$, \begin{align*} \kappa(\kappa(x,y)t_\alpha,h) &= \kappa(x,y)\kappa(t_\alpha,h) \\ &= \kappa(x,y)\alpha(h). \end{align*} Therefore \begin{align*} \kappa([x,y]-\kappa(x,y)t_\alpha,h)=0 \end{align*} for every $h \in \mathfrak{h}$.[/step]

[guided]We now determine the element $[x,y] \in \mathfrak{h}$ by testing it against arbitrary elements of $\mathfrak{h}$. Fix $h \in \mathfrak{h}$. The Killing form is invariant, meaning that for all $a,b,c \in \mathfrak{g}$, \begin{align*} \kappa([a,b],c)=\kappa(a,[b,c]). \end{align*} Applying this with $a=x$, $b=y$, and $c=h$ gives \begin{align*} \kappa([x,y],h)=\kappa(x,[y,h]). \end{align*} Because $y \in \mathfrak{g}_{-\alpha}$, the defining relation for the root space says \begin{align*} [h,y] = -\alpha(h)y. \end{align*} Antisymmetry of the Lie bracket then gives \begin{align*} [y,h] = \alpha(h)y. \end{align*} Substituting this into the invariant-form identity gives \begin{align*} \kappa([x,y],h) &= \kappa(x,\alpha(h)y) \\ &= \alpha(h)\kappa(x,y). \end{align*} Now compare this with the element $\kappa(x,y)t_\alpha \in \mathfrak{h}$. By definition, $t_\alpha$ is the unique element of $\mathfrak{h}$ satisfying \begin{align*} \kappa(t_\alpha,h)=\alpha(h) \end{align*} for every $h \in \mathfrak{h}$. Hence \begin{align*} \kappa(\kappa(x,y)t_\alpha,h) &= \kappa(x,y)\kappa(t_\alpha,h) \\ &= \kappa(x,y)\alpha(h). \end{align*} Thus $[x,y]$ and $\kappa(x,y)t_\alpha$ have the same Killing-form pairing with every $h \in \mathfrak{h}$, so \begin{align*} \kappa([x,y]-\kappa(x,y)t_\alpha,h)=0 \end{align*} for every $h \in \mathfrak{h}$.[/guided]

[step:Use nondegeneracy on $\mathfrak{h}$ to conclude the bracket formula] The element \begin{align*} w := [x,y]-\kappa(x,y)t_\alpha \end{align*} belongs to $\mathfrak{h}$ because both $[x,y]$ and $t_\alpha$ belong to $\mathfrak{h}$. The previous step shows that \begin{align*} \kappa(w,h)=0 \end{align*} for every $h \in \mathfrak{h}$. Since the restriction of $\kappa$ to $\mathfrak{h} \times \mathfrak{h}$ is nondegenerate in the semisimple Cartan decomposition setting, we obtain $w=0$. Therefore \begin{align*} [x,y]=\kappa(x,y)t_\alpha. \end{align*} This proves both assertions. [/step]