[proofplan] We use that the adjoint action of a Lie group element is a [Lie algebra](/page/Lie%20Algebra) automorphism. This gives an intertwining identity between the infinitesimal adjoint maps $\operatorname{ad}_X$ and $\operatorname{ad}_{\operatorname{Ad}_gX}$. Substituting this identity into the trace formula defining the Killing form turns $B(\operatorname{Ad}_gX,\operatorname{Ad}_gY)$ into the trace of a conjugate of $\operatorname{ad}_X\circ\operatorname{ad}_Y$, and conjugation invariance of trace gives the desired equality. [/proofplan]

[step:Intertwine the infinitesimal adjoint maps by the adjoint representation]Fix $g\in G$. Define the conjugation map \begin{align*} C_g:G\to G,\qquad h\mapsto ghg^{-1}. \end{align*} By the definition of the adjoint representation [citetheorem:8821], $\operatorname{Ad}_g=d(C_g)_e:\mathfrak g\to\mathfrak g$. Since $C_g$ is a Lie group automorphism with inverse $C_{g^{-1}}$, its differential $\operatorname{Ad}_g$ is a linear isomorphism with inverse $\operatorname{Ad}_{g^{-1}}$. The [differential of a Lie group homomorphism preserves Lie brackets](/theorems/8803) [citetheorem:8803]. Applying this to the automorphism $C_g:G\to G$, we obtain, for all $U,V\in\mathfrak g$, \begin{align*} \operatorname{Ad}_g[U,V]=[\operatorname{Ad}_gU,\operatorname{Ad}_gV]. \end{align*} Let $X\in\mathfrak g$. We claim that \begin{align*} \operatorname{ad}_{\operatorname{Ad}_gX}=\operatorname{Ad}_g\circ\operatorname{ad}_X\circ\operatorname{Ad}_{g^{-1}}. \end{align*} Indeed, for every $Z\in\mathfrak g$, \begin{align*} (\operatorname{Ad}_g\circ\operatorname{ad}_X\circ\operatorname{Ad}_{g^{-1}})(Z)=\operatorname{Ad}_g[X,\operatorname{Ad}_{g^{-1}}Z]. \end{align*} Using bracket preservation with $U=X$ and $V=\operatorname{Ad}_{g^{-1}}Z$ gives \begin{align*} \operatorname{Ad}_g[X,\operatorname{Ad}_{g^{-1}}Z]=[\operatorname{Ad}_gX,\operatorname{Ad}_g\operatorname{Ad}_{g^{-1}}Z]. \end{align*} Since $\operatorname{Ad}_{g^{-1}}$ is the inverse of $\operatorname{Ad}_g$, this equals \begin{align*} [\operatorname{Ad}_gX,Z]=\operatorname{ad}_{\operatorname{Ad}_gX}(Z). \end{align*} Thus the two linear maps agree on every $Z\in\mathfrak g$.[/step]

[guided]Fix $g\in G$. The point of this step is to translate the group-level conjugation action into a statement about the linear maps whose traces define the Killing form. Define \begin{align*} C_g:G\to G,\qquad h\mapsto ghg^{-1}. \end{align*} This is a smooth Lie group automorphism, with inverse $C_{g^{-1}}:G\to G$. By the definition of the adjoint representation [citetheorem:8821], its differential at the identity is \begin{align*} \operatorname{Ad}_g=d(C_g)_e:\mathfrak g\to\mathfrak g. \end{align*} Since $C_{g^{-1}}$ is the inverse of $C_g$, the chain rule gives that $\operatorname{Ad}_{g^{-1}}$ is the inverse [linear map](/page/Linear%20Map) of $\operatorname{Ad}_g$. We now use the fact that differentials of Lie group homomorphisms preserve Lie brackets [citetheorem:8803]. The map $C_g:G\to G$ is a Lie [group homomorphism](/page/Group%20Homomorphism) because, for all $a,b\in G$, \begin{align*} C_g(ab)=gabg^{-1}=(gag^{-1})(gbg^{-1})=C_g(a)C_g(b). \end{align*} Therefore its differential preserves brackets: for all $U,V\in\mathfrak g$, \begin{align*} \operatorname{Ad}_g[U,V]=[\operatorname{Ad}_gU,\operatorname{Ad}_gV]. \end{align*} We want an identity involving $\operatorname{ad}_{\operatorname{Ad}_gX}$, which is the map that brackets with $\operatorname{Ad}_gX$. Let $X\in\mathfrak g$ and $Z\in\mathfrak g$. Because $\operatorname{Ad}_{g^{-1}}$ is the inverse of $\operatorname{Ad}_g$, we may first pull $Z$ back to $\operatorname{Ad}_{g^{-1}}Z$, bracket with $X$, and then push forward by $\operatorname{Ad}_g$: \begin{align*} (\operatorname{Ad}_g\circ\operatorname{ad}_X\circ\operatorname{Ad}_{g^{-1}})(Z)=\operatorname{Ad}_g[X,\operatorname{Ad}_{g^{-1}}Z]. \end{align*} Applying bracket preservation with $U=X$ and $V=\operatorname{Ad}_{g^{-1}}Z$ gives \begin{align*} \operatorname{Ad}_g[X,\operatorname{Ad}_{g^{-1}}Z]=[\operatorname{Ad}_gX,\operatorname{Ad}_g\operatorname{Ad}_{g^{-1}}Z]. \end{align*} The last factor simplifies because $\operatorname{Ad}_g\operatorname{Ad}_{g^{-1}}=\operatorname{id}_{\mathfrak g}$, so \begin{align*} [\operatorname{Ad}_gX,\operatorname{Ad}_g\operatorname{Ad}_{g^{-1}}Z]=[\operatorname{Ad}_gX,Z]. \end{align*} By the definition of $\operatorname{ad}_{\operatorname{Ad}_gX}$, this final expression is \begin{align*} [\operatorname{Ad}_gX,Z]=\operatorname{ad}_{\operatorname{Ad}_gX}(Z). \end{align*} Since the equality holds for every $Z\in\mathfrak g$, we have proved the linear-map identity \begin{align*} \operatorname{ad}_{\operatorname{Ad}_gX}=\operatorname{Ad}_g\circ\operatorname{ad}_X\circ\operatorname{Ad}_{g^{-1}}. \end{align*}[/guided]

[step:Substitute the intertwining identities into the trace defining the Killing form] Let $X,Y\in\mathfrak g$. By the definition of the Killing form, \begin{align*} B(\operatorname{Ad}_gX,\operatorname{Ad}_gY)=\operatorname{tr}(\operatorname{ad}_{\operatorname{Ad}_gX}\circ\operatorname{ad}_{\operatorname{Ad}_gY}). \end{align*} Applying the intertwining identity from the previous step to both $X$ and $Y$ gives \begin{align*} \operatorname{ad}_{\operatorname{Ad}_gX}\circ\operatorname{ad}_{\operatorname{Ad}_gY}=(\operatorname{Ad}_g\circ\operatorname{ad}_X\circ\operatorname{Ad}_{g^{-1}})\circ(\operatorname{Ad}_g\circ\operatorname{ad}_Y\circ\operatorname{Ad}_{g^{-1}}). \end{align*} Using associativity of composition and $\operatorname{Ad}_{g^{-1}}\circ\operatorname{Ad}_g=\operatorname{id}_{\mathfrak g}$, this composition is \begin{align*} \operatorname{Ad}_g\circ\operatorname{ad}_X\circ\operatorname{ad}_Y\circ\operatorname{Ad}_{g^{-1}}. \end{align*} Hence \begin{align*} B(\operatorname{Ad}_gX,\operatorname{Ad}_gY)=\operatorname{tr}(\operatorname{Ad}_g\circ\operatorname{ad}_X\circ\operatorname{ad}_Y\circ\operatorname{Ad}_{g^{-1}}). \end{align*} [/step]

[step:Use conjugation invariance of trace to recover the original Killing form] Set \begin{align*} T:=\operatorname{ad}_X\circ\operatorname{ad}_Y:\mathfrak g\to\mathfrak g. \end{align*} Since $\mathfrak g$ is finite-dimensional and $\operatorname{Ad}_g:\mathfrak g\to\mathfrak g$ is an invertible linear map with inverse $\operatorname{Ad}_{g^{-1}}$, the elementary trace identity \begin{align*} \operatorname{tr}(STS^{-1})=\operatorname{tr}(T) \end{align*} applies with $S=\operatorname{Ad}_g$; this follows from cyclicity of trace, because $\operatorname{tr}(STS^{-1})=\operatorname{tr}(TS^{-1}S)=\operatorname{tr}(T)$. Therefore \begin{align*} \operatorname{tr}(\operatorname{Ad}_g\circ T\circ\operatorname{Ad}_{g^{-1}})=\operatorname{tr}(T). \end{align*} Substituting the definition of $T$ gives \begin{align*} B(\operatorname{Ad}_gX,\operatorname{Ad}_gY)=\operatorname{tr}(\operatorname{ad}_X\circ\operatorname{ad}_Y)=B(X,Y). \end{align*} Since $g\in G$ and $X,Y\in\mathfrak g$ were arbitrary, the Killing form is invariant under the adjoint action of $G$. [/step]