[proofplan] We prove the identity by translating the Lie bracket into commutators of adjoint endomorphisms. The Jacobi identity gives $\operatorname{ad}_{[x,y]} = \operatorname{ad}_x \circ \operatorname{ad}_y - \operatorname{ad}_y \circ \operatorname{ad}_x$. After substituting this into the definition of the Killing form, the result follows from the cyclicity of the trace for finite-dimensional endomorphisms. [/proofplan] [step:Rewrite the adjoint of a bracket as a commutator of adjoint maps] Fix $x,y,z \in \mathfrak g$. For every $u \in \mathfrak g$, the Jacobi identity gives \begin{align*} [[x,y],u] = [x,[y,u]] - [y,[x,u]]. \end{align*} By the definition of the adjoint endomorphisms, this says \begin{align*} \operatorname{ad}_{[x,y]}(u) = (\operatorname{ad}_x \circ \operatorname{ad}_y)(u) - (\operatorname{ad}_y \circ \operatorname{ad}_x)(u). \end{align*} Since this holds for every $u \in \mathfrak g$, \begin{align*} \operatorname{ad}_{[x,y]} = \operatorname{ad}_x \circ \operatorname{ad}_y - \operatorname{ad}_y \circ \operatorname{ad}_x. \end{align*} [/step] [step:Record cyclicity of trace for finite-dimensional endomorphisms] Let $V$ be a finite-dimensional [vector space](/page/Vector%20Space) over $k$. For any $A,B \in \operatorname{End}_k(V)$, \begin{align*} \operatorname{tr}(A \circ B)=\operatorname{tr}(B \circ A). \end{align*} Indeed, choosing a basis of $V$ and writing the matrices of $A$ and $B$ as $(A_{ij})$ and $(B_{ij})$, respectively, we compute \begin{align*} \operatorname{tr}(AB) &= \sum_i (AB)_{ii} \\ &= \sum_i \sum_j A_{ij}B_{ji} \\ &= \sum_j \sum_i B_{ji}A_{ij} \\ &= \sum_j (BA)_{jj} \\ &= \operatorname{tr}(BA). \end{align*} Thus the trace is cyclic for products of two endomorphisms, and therefore also allows cyclic rotation inside longer finite products, for example \begin{align*} \operatorname{tr}(A \circ B \circ C) = \operatorname{tr}(B \circ C \circ A) = \operatorname{tr}(C \circ A \circ B). \end{align*} [/step] [step:Apply cyclicity to the trace expression for the Killing form] Using the definition of $\kappa$ and the commutator identity from the first step, we obtain \begin{align*} \kappa([x,y],z) &= \operatorname{tr}(\operatorname{ad}_{[x,y]} \circ \operatorname{ad}_z) \\ &= \operatorname{tr}\left((\operatorname{ad}_x \circ \operatorname{ad}_y - \operatorname{ad}_y \circ \operatorname{ad}_x) \circ \operatorname{ad}_z\right) \\ &= \operatorname{tr}(\operatorname{ad}_x \circ \operatorname{ad}_y \circ \operatorname{ad}_z) - \operatorname{tr}(\operatorname{ad}_y \circ \operatorname{ad}_x \circ \operatorname{ad}_z). \end{align*} By cyclicity of trace applied to the second term, \begin{align*} \operatorname{tr}(\operatorname{ad}_y \circ \operatorname{ad}_x \circ \operatorname{ad}_z) = \operatorname{tr}(\operatorname{ad}_x \circ \operatorname{ad}_z \circ \operatorname{ad}_y). \end{align*} Therefore \begin{align*} \kappa([x,y],z) &= \operatorname{tr}(\operatorname{ad}_x \circ \operatorname{ad}_y \circ \operatorname{ad}_z) - \operatorname{tr}(\operatorname{ad}_x \circ \operatorname{ad}_z \circ \operatorname{ad}_y) \\ &= \operatorname{tr}\left(\operatorname{ad}_x \circ (\operatorname{ad}_y \circ \operatorname{ad}_z - \operatorname{ad}_z \circ \operatorname{ad}_y)\right). \end{align*} [guided] The goal is to transform the left-hand side into an expression involving $\operatorname{ad}_{[y,z]}$. From the first step, the adjoint representation converts Lie brackets into commutators of endomorphisms: \begin{align*} \operatorname{ad}_{[x,y]} = \operatorname{ad}_x \circ \operatorname{ad}_y - \operatorname{ad}_y \circ \operatorname{ad}_x. \end{align*} Substituting this into the Killing form gives \begin{align*} \kappa([x,y],z) &= \operatorname{tr}(\operatorname{ad}_{[x,y]} \circ \operatorname{ad}_z) \\ &= \operatorname{tr}\left((\operatorname{ad}_x \circ \operatorname{ad}_y - \operatorname{ad}_y \circ \operatorname{ad}_x) \circ \operatorname{ad}_z\right) \\ &= \operatorname{tr}(\operatorname{ad}_x \circ \operatorname{ad}_y \circ \operatorname{ad}_z) - \operatorname{tr}(\operatorname{ad}_y \circ \operatorname{ad}_x \circ \operatorname{ad}_z). \end{align*} The second trace has the factors in the wrong order. Since $\mathfrak g$ is finite-dimensional, trace is cyclic for endomorphisms of $\mathfrak g$, so \begin{align*} \operatorname{tr}(\operatorname{ad}_y \circ \operatorname{ad}_x \circ \operatorname{ad}_z) = \operatorname{tr}(\operatorname{ad}_x \circ \operatorname{ad}_z \circ \operatorname{ad}_y). \end{align*} Now both terms begin with $\operatorname{ad}_x$, so they can be combined: \begin{align*} \kappa([x,y],z) &= \operatorname{tr}(\operatorname{ad}_x \circ \operatorname{ad}_y \circ \operatorname{ad}_z) - \operatorname{tr}(\operatorname{ad}_x \circ \operatorname{ad}_z \circ \operatorname{ad}_y) \\ &= \operatorname{tr}\left(\operatorname{ad}_x \circ (\operatorname{ad}_y \circ \operatorname{ad}_z - \operatorname{ad}_z \circ \operatorname{ad}_y)\right). \end{align*} This is exactly the form needed, because the parenthesized commutator is $\operatorname{ad}_{[y,z]}$. [/guided] [/step] [step:Identify the remaining commutator and conclude invariance] Applying the commutator identity from the first step with $y$ and $z$ in place of $x$ and $y$ gives \begin{align*} \operatorname{ad}_{[y,z]} = \operatorname{ad}_y \circ \operatorname{ad}_z - \operatorname{ad}_z \circ \operatorname{ad}_y. \end{align*} Substituting this into the final trace expression from the previous step yields \begin{align*} \kappa([x,y],z) &= \operatorname{tr}(\operatorname{ad}_x \circ \operatorname{ad}_{[y,z]}) \\ &= \kappa(x,[y,z]). \end{align*} Since $x,y,z \in \mathfrak g$ were arbitrary, the Killing form is invariant: \begin{align*} \kappa([x,y],z)=\kappa(x,[y,z]) \end{align*} for all $x,y,z \in \mathfrak g$. [/step]