[proofplan] Fix a root $\alpha$ and choose a non-zero vector in the root space $\mathfrak{g}_{\alpha}$. The Killing form pairs $\mathfrak{g}_{\alpha}$ nondegenerately with $\mathfrak{g}_{-\alpha}$, so we choose an opposite-root vector with non-zero pairing. Invariance of the Killing form identifies their bracket as a non-zero scalar multiple of $t_{\alpha}$, and rescaling makes the bracket equal to $h_{\alpha}$. The root-space definitions and the normalization of $h_{\alpha}$ then give the two remaining $\mathfrak{sl}_2$ bracket relations. [/proofplan]

[step:Choose opposite-root vectors with non-zero Killing pairing] Fix $\alpha\in\Phi$. Since $\alpha$ is a root, the root space $\mathfrak{g}_{\alpha}$ is non-zero. Choose \begin{align*} x\in\mathfrak{g}_{\alpha}\setminus\{0\}. \end{align*} By the [Orthogonality Of Root Spaces](/theorems/TEMP-12) and the [Pairing Of Opposite Root Spaces](/theorems/TEMP-13) for the root-space decomposition of a semisimple Lie algebra, the restriction of the Killing form to \begin{align*} \mathfrak{g}_{\alpha}\times \mathfrak{g}_{-\alpha} \end{align*} is nondegenerate. The hypotheses apply because $\mathfrak{g}$ is semisimple, $\mathfrak{h}$ is the Cartan subalgebra used to define $\Phi$, and $\alpha\in\Phi$. Hence there exists \begin{align*} y\in\mathfrak{g}_{-\alpha} \end{align*} such that $\kappa(x,y)\ne 0$. [/step]

[step:Compute the bracket $[x,y]$ by Killing form invariance] We first show that $[x,y]=\kappa(x,y)t_{\alpha}$. Define the zero-weight space for the adjoint action of $\mathfrak{h}$ by \begin{align*} \mathfrak{g}_{0}:=\{z\in\mathfrak{g}: [h,z]=0\text{ for every }h\in\mathfrak{h}\}. \end{align*} By the [Root Space Bracket Rule](/theorems/TEMP-11), \begin{align*} [\mathfrak{g}_{\alpha},\mathfrak{g}_{-\alpha}]\subseteq \mathfrak{g}_{\alpha+(-\alpha)}=\mathfrak{g}_{0}, \end{align*} and by the [Root Space Decomposition](/theorems/TEMP-10) attached to $\mathfrak{h}$ one has $\mathfrak{g}_{0}=\mathfrak{h}$. These hypotheses apply because $\alpha$ and $-\alpha$ are roots and $\mathfrak{h}$ is the Cartan subalgebra defining the root spaces. Thus $[x,y]\in\mathfrak{h}$. For every $h\in\mathfrak{h}$, the [Invariance of the Killing Form](/theorems/3808) and symmetry of the Killing form give \begin{align*} \kappa([x,y],h) &=\kappa(x,[y,h])\\ &=\kappa(x,\alpha(h)y)\\ &=\alpha(h)\kappa(x,y)\\ &=\kappa(\kappa(x,y)t_{\alpha},h). \end{align*} The restriction $\kappa|_{\mathfrak{h}\times\mathfrak{h}}$ is nondegenerate by the Cartan-subalgebra nondegeneracy consequence of the [Root Space Decomposition](/theorems/TEMP-10) and the [Cartan Semisimplicity Criterion](/theorems/3813); its hypotheses apply because $\mathfrak{g}$ is semisimple and $\mathfrak{h}$ is the Cartan subalgebra in the root-space decomposition. Therefore equality of pairings against every $h\in\mathfrak{h}$ implies \begin{align*} [x,y]=\kappa(x,y)t_{\alpha}. \end{align*}[/step]

[guided]Recall the data chosen in the previous step: $x\in\mathfrak{g}_{\alpha}\setminus\{0\}$ and $y\in\mathfrak{g}_{-\alpha}$ with $\kappa(x,y)\ne 0$. We want to identify the bracket $[x,y]$ as an element of $\mathfrak{h}$. Define the zero-weight space by \begin{align*} \mathfrak{g}_{0}:=\{z\in\mathfrak{g}: [h,z]=0\text{ for every }h\in\mathfrak{h}\}. \end{align*} Because $x\in\mathfrak{g}_{\alpha}$ and $y\in\mathfrak{g}_{-\alpha}$, the [Root Space Bracket Rule](/theorems/TEMP-11) applies to the pair of roots $\alpha$ and $-\alpha$ and gives \begin{align*} [x,y]\in \mathfrak{g}_{\alpha+(-\alpha)}=\mathfrak{g}_{0}. \end{align*} For the [Root Space Decomposition](/theorems/TEMP-10) attached to the Cartan subalgebra $\mathfrak{h}$, the zero-weight space is exactly $\mathfrak{h}$. Thus $[x,y]\in\mathfrak{h}$. To determine which element of $\mathfrak{h}$ it is, pair it with an arbitrary $h\in\mathfrak{h}$. The [Invariance of the Killing Form](/theorems/3808) says that \begin{align*} \kappa([a,b],c)=\kappa(a,[b,c]) \end{align*} for all $a,b,c\in\mathfrak{g}$. Applying this with $a=x$, $b=y$, and $c=h$, we obtain \begin{align*} \kappa([x,y],h)=\kappa(x,[y,h]). \end{align*} Since $y\in\mathfrak{g}_{-\alpha}$, we have $[h,y]=-\alpha(h)y$, hence $[y,h]=\alpha(h)y$. Therefore \begin{align*} \kappa([x,y],h) &=\kappa(x,\alpha(h)y)\\ &=\alpha(h)\kappa(x,y). \end{align*} By definition of $t_{\alpha}$, $\kappa(t_{\alpha},h)=\alpha(h)$ for every $h\in\mathfrak{h}$, so \begin{align*} \alpha(h)\kappa(x,y)=\kappa(\kappa(x,y)t_{\alpha},h). \end{align*} Thus $[x,y]$ and $\kappa(x,y)t_{\alpha}$ have the same Killing pairing with every element of $\mathfrak{h}$. The Cartan-subalgebra nondegeneracy consequence of the [Root Space Decomposition](/theorems/TEMP-10) and the [Cartan Semisimplicity Criterion](/theorems/3813) says that $\kappa|_{\mathfrak{h}\times\mathfrak{h}}$ is nondegenerate; its hypotheses apply here because $\mathfrak{g}$ is semisimple and $\mathfrak{h}$ is precisely the Cartan subalgebra used in the root-space decomposition. Hence the difference $[x,y]-\kappa(x,y)t_{\alpha}\in\mathfrak{h}$ pairs to zero with all of $\mathfrak{h}$, so the difference is zero: \begin{align*} [x,y]=\kappa(x,y)t_{\alpha}. \end{align*}[/guided]

[step:Rescale the vectors so their bracket is $h_{\alpha}$] We have $\kappa(x,y)\ne 0$ by construction. Also $\kappa(t_{\alpha},t_{\alpha})\ne 0$ by the root non-isotropy consequence of the [Pairing Of Opposite Root Spaces](/theorems/TEMP-13); equivalently, the coroot element $h_{\alpha}=2t_{\alpha}/\kappa(t_{\alpha},t_{\alpha})$ is well-defined. Therefore the following rescaling is legitimate. Define \begin{align*} e_{\alpha}:=\frac{2}{\kappa(x,y)\kappa(t_{\alpha},t_{\alpha})}x, \qquad f_{\alpha}:=y. \end{align*} Since $\mathfrak{g}_{\alpha}$ and $\mathfrak{g}_{-\alpha}$ are vector spaces, $e_{\alpha}\in\mathfrak{g}_{\alpha}$ and $f_{\alpha}\in\mathfrak{g}_{-\alpha}$. Using bilinearity of the Lie bracket and the bracket computation above, \begin{align*} [e_{\alpha},f_{\alpha}] &=\frac{2}{\kappa(x,y)\kappa(t_{\alpha},t_{\alpha})}[x,y]\\ &=\frac{2}{\kappa(x,y)\kappa(t_{\alpha},t_{\alpha})}\kappa(x,y)t_{\alpha}\\ &=\frac{2t_{\alpha}}{\kappa(t_{\alpha},t_{\alpha})}\\ &=h_{\alpha}. \end{align*} [/step]

[step:Use the root-space action of $h_{\alpha}$ to obtain the remaining brackets]Because $e_{\alpha}\in\mathfrak{g}_{\alpha}$, the definition of $\mathfrak{g}_{\alpha}$ gives \begin{align*} [h_{\alpha},e_{\alpha}]=\alpha(h_{\alpha})e_{\alpha}. \end{align*} By the definition of $h_{\alpha}$ and $t_{\alpha}$, \begin{align*} \alpha(h_{\alpha}) &=\kappa(t_{\alpha},h_{\alpha})\\ &=\kappa\left(t_{\alpha},\frac{2t_{\alpha}}{\kappa(t_{\alpha},t_{\alpha})}\right)\\ &=2. \end{align*} Hence \begin{align*} [h_{\alpha},e_{\alpha}]=2e_{\alpha}. \end{align*} Similarly, since $f_{\alpha}\in\mathfrak{g}_{-\alpha}$, \begin{align*} [h_{\alpha},f_{\alpha}] =(-\alpha)(h_{\alpha})f_{\alpha} =-2f_{\alpha}. \end{align*}[/step]

[guided]The remaining two brackets come directly from the defining property of root spaces. Since $e_{\alpha}\in\mathfrak{g}_{\alpha}$, every $h\in\mathfrak{h}$ acts on $e_{\alpha}$ by the scalar $\alpha(h)$: \begin{align*} [h,e_{\alpha}]=\alpha(h)e_{\alpha}. \end{align*} We apply this with $h=h_{\alpha}$. This gives \begin{align*} [h_{\alpha},e_{\alpha}]=\alpha(h_{\alpha})e_{\alpha}. \end{align*} Now compute the scalar. By definition, $t_{\alpha}$ represents the functional $\alpha$ under the Killing form on $\mathfrak{h}$, so $\alpha(h)=\kappa(t_{\alpha},h)$ for every $h\in\mathfrak{h}$. Therefore \begin{align*} \alpha(h_{\alpha}) &=\kappa(t_{\alpha},h_{\alpha})\\ &=\kappa\left(t_{\alpha},\frac{2t_{\alpha}}{\kappa(t_{\alpha},t_{\alpha})}\right)\\ &=2. \end{align*} Thus \begin{align*} [h_{\alpha},e_{\alpha}]=2e_{\alpha}. \end{align*} The same argument applies to the opposite root. Since $f_{\alpha}\in\mathfrak{g}_{-\alpha}$, every $h\in\mathfrak{h}$ acts on $f_{\alpha}$ by the scalar $-\alpha(h)$. Taking again $h=h_{\alpha}$ gives \begin{align*} [h_{\alpha},f_{\alpha}] =(-\alpha)(h_{\alpha})f_{\alpha} =-2f_{\alpha}. \end{align*}[/guided]

[step:Identify the generated subalgebra with $\mathfrak{sl}_2(k)$] Let \begin{align*} \mathfrak{s}_{\alpha}:=\operatorname{span}_k\{e_{\alpha},h_{\alpha},f_{\alpha}\}\subseteq\mathfrak{g}. \end{align*} The three bracket identities \begin{align*} [e_{\alpha},f_{\alpha}]=h_{\alpha}, \qquad [h_{\alpha},e_{\alpha}]=2e_{\alpha}, \qquad [h_{\alpha},f_{\alpha}]=-2f_{\alpha} \end{align*} show that $\mathfrak{s}_{\alpha}$ is closed under the Lie bracket, hence is a Lie subalgebra of $\mathfrak{g}$. Define a $k$-[linear map](/page/Linear%20Map) \begin{align*} \varphi:\mathfrak{sl}_2(k)&\to\mathfrak{s}_{\alpha}\\ \begin{pmatrix}0&1\\0&0\end{pmatrix}&\mapsto e_{\alpha}\\ \begin{pmatrix}1&0\\0&-1\end{pmatrix}&\mapsto h_{\alpha}\\ \begin{pmatrix}0&0\\1&0\end{pmatrix}&\mapsto f_{\alpha}. \end{align*} The standard basis of $\mathfrak{sl}_2(k)$ satisfies \begin{align*} [h,e]=2e, \qquad [h,f]=-2f, \qquad [e,f]=h. \end{align*} The defining bracket relations for $e_{\alpha},h_{\alpha},f_{\alpha}$ therefore imply that $\varphi$ preserves brackets. It remains to check bijectivity. The map $\varphi$ is surjective by the definition of $\mathfrak{s}_{\alpha}$ as the span of $e_{\alpha},h_{\alpha},f_{\alpha}$. For injectivity, suppose $a,b,c\in k$ satisfy \begin{align*} ae_{\alpha}+bh_{\alpha}+cf_{\alpha}=0. \end{align*} Here $e_{\alpha}\in\mathfrak{g}_{\alpha}$ is non-zero because it is a non-zero scalar multiple of $x$, $f_{\alpha}\in\mathfrak{g}_{-\alpha}$ is non-zero because $\kappa(x,y)\ne0$, and $h_{\alpha}\in\mathfrak{h}=\mathfrak{g}_{0}$ is non-zero because $\kappa(t_{\alpha},t_{\alpha})\ne0$. The root-space decomposition is direct, so the three summands lie in the direct sum \begin{align*} \mathfrak{g}_{\alpha}\oplus\mathfrak{g}_{0}\oplus\mathfrak{g}_{-\alpha}. \end{align*} Therefore $ae_{\alpha}=0$, $bh_{\alpha}=0$, and $cf_{\alpha}=0$, hence $a=b=c=0$. Thus $e_{\alpha},h_{\alpha},f_{\alpha}$ are linearly independent, so $\varphi$ is injective. Hence $\varphi$ is a Lie algebra isomorphism, and \begin{align*} \mathfrak{s}_{\alpha}\cong \mathfrak{sl}_2(k). \end{align*} This proves the theorem. [/step]