[proofplan] We prove both implications by relating the radical of the Killing form to solvable ideals. First, if the Killing form has a nonzero kernel, [invariance of the Killing form](/theorems/3808) makes that kernel an ideal, and Cartan's solvability criterion shows that it is solvable. Thus a semisimple Lie algebra has nondegenerate Killing form. Conversely, if $\mathfrak g$ has a nonzero solvable radical, the last nonzero derived ideal of the radical is a nonzero abelian ideal, and every element of that abelian ideal lies in the kernel of the Killing form. Hence nondegeneracy forces the radical to vanish. [/proofplan]

[step:Define the kernel of the Killing form and prove it is an ideal]Define the kernel of the Killing form as the subspace \begin{align*} \mathfrak k := \{x \in \mathfrak g : \kappa_{\mathfrak g}(x,y)=0 \text{ for all } y \in \mathfrak g\}. \end{align*} We first show that $\mathfrak k \trianglelefteq \mathfrak g$. Let $x \in \mathfrak k$, let $z \in \mathfrak g$, and let $y \in \mathfrak g$. The Killing form is invariant, meaning \begin{align*} \kappa_{\mathfrak g}([z,x],y)=\kappa_{\mathfrak g}(z,[x,y]). \end{align*} Equivalently, using symmetry and invariance, \begin{align*} \kappa_{\mathfrak g}([z,x],y) = \kappa_{\mathfrak g}(x,[y,z]). \end{align*} Since $x \in \mathfrak k$ and $[y,z]\in \mathfrak g$, the last expression is $0$. Therefore \begin{align*} \kappa_{\mathfrak g}([z,x],y)=0 \end{align*} for every $y \in \mathfrak g$, so $[z,x]\in \mathfrak k$. Hence $\mathfrak k$ is an ideal of $\mathfrak g$.[/step]

[guided]We want to turn degeneracy of the Killing form into an algebraic obstruction inside $\mathfrak g$. The relevant object is the full kernel of the [bilinear form](/page/Bilinear%20Form): \begin{align*} \mathfrak k := \{x \in \mathfrak g : \kappa_{\mathfrak g}(x,y)=0 \text{ for all } y \in \mathfrak g\}. \end{align*} This is a vector subspace because $\kappa_{\mathfrak g}$ is bilinear. To use semisimplicity, however, we need $\mathfrak k$ to be an ideal, not merely a subspace. Let $x \in \mathfrak k$, let $z \in \mathfrak g$, and let $y \in \mathfrak g$. We must prove that $[z,x]\in \mathfrak k$, meaning that its Killing pairing with every $y\in\mathfrak g$ is zero. The Killing form is invariant under the adjoint action: \begin{align*} \kappa_{\mathfrak g}([z,x],y)=\kappa_{\mathfrak g}(z,[x,y]). \end{align*} Using symmetry of $\kappa_{\mathfrak g}$ and the same invariance identity, this can also be written as \begin{align*} \kappa_{\mathfrak g}([z,x],y) = \kappa_{\mathfrak g}(x,[y,z]). \end{align*} Now $[y,z]\in \mathfrak g$, and $x$ pairs to zero with every element of $\mathfrak g$ by the definition of $\mathfrak k$. Therefore \begin{align*} \kappa_{\mathfrak g}([z,x],y)=0. \end{align*} Since this holds for every $y\in\mathfrak g$, we have $[z,x]\in\mathfrak k$. Thus $\mathfrak k$ is closed under brackets with arbitrary elements of $\mathfrak g$, so $\mathfrak k \trianglelefteq \mathfrak g$.[/guided]

[step:Use Cartan solvability criterion to prove semisimplicity implies nondegeneracy]Assume that $\mathfrak g$ is semisimple. Since $\mathfrak k$ is an ideal, its derived algebra $[\mathfrak k,\mathfrak k]$ is contained in $\mathfrak k$. Hence, for every $x\in[\mathfrak k,\mathfrak k]$ and every $y\in\mathfrak g$, \begin{align*} \operatorname{tr}(\operatorname{ad}_x\circ \operatorname{ad}_y) = \kappa_{\mathfrak g}(x,y) =0. \end{align*} By Cartan's solvability criterion for ideals (citing a result not yet in the wiki: Cartan's Solvability Criterion), the ideal $\mathfrak k$ is solvable. Because $\mathfrak g$ is semisimple, its solvable radical is $0$, so every solvable ideal of $\mathfrak g$ is zero. Therefore $\mathfrak k=0$. The kernel of $\kappa_{\mathfrak g}$ is zero, so $\kappa_{\mathfrak g}$ is nondegenerate.[/step]

[guided]Now assume that $\mathfrak g$ is semisimple. From the previous step, the kernel \begin{align*} \mathfrak k := \{x \in \mathfrak g : \kappa_{\mathfrak g}(x,y)=0 \text{ for all } y \in \mathfrak g\} \end{align*} is an ideal of $\mathfrak g$. To prove that $\kappa_{\mathfrak g}$ is nondegenerate, we must prove $\mathfrak k=0$. The key tool is Cartan's solvability criterion for ideals (citing a result not yet in the wiki: Cartan's Solvability Criterion). In the form used here, it says: if $\mathfrak a\trianglelefteq\mathfrak g$ satisfies \begin{align*} \operatorname{tr}(\operatorname{ad}_x\circ \operatorname{ad}_y)=0 \end{align*} for every $x\in[\mathfrak a,\mathfrak a]$ and every $y\in\mathfrak g$, then $\mathfrak a$ is solvable. We apply this criterion with $\mathfrak a=\mathfrak k$. Since $\mathfrak k$ is an ideal, $[\mathfrak k,\mathfrak k]\subseteq \mathfrak k$. Therefore, if $x\in[\mathfrak k,\mathfrak k]$ and $y\in\mathfrak g$, then $x\in\mathfrak k$, and the definition of $\mathfrak k$ gives \begin{align*} \operatorname{tr}(\operatorname{ad}_x\circ \operatorname{ad}_y) = \kappa_{\mathfrak g}(x,y) =0. \end{align*} All hypotheses of Cartan's solvability criterion are therefore satisfied, so $\mathfrak k$ is solvable. Semisimplicity of $\mathfrak g$ means that the solvable radical of $\mathfrak g$, namely the largest solvable ideal of $\mathfrak g$, is zero. Since $\mathfrak k$ is a solvable ideal, it must be contained in that radical. Hence $\mathfrak k=0$. Thus the Killing form has zero kernel, which is precisely nondegeneracy.[/guided]

[step:Extract a nonzero abelian ideal from a nonzero solvable radical]Conversely, assume that $\kappa_{\mathfrak g}$ is nondegenerate. Let $\mathfrak r:=\operatorname{rad}(\mathfrak g)$ be the solvable radical of $\mathfrak g$. We prove that $\mathfrak r=0$. Suppose, for contradiction, that $\mathfrak r\neq 0$. Define the derived series of $\mathfrak r$ by \begin{align*} \mathfrak r_0 &:= \mathfrak r,\\ \mathfrak r_{m+1} &:= [\mathfrak r_m,\mathfrak r_m]\quad \text{for } m\geq 0. \end{align*} Since $\mathfrak r$ is solvable, there exists $N\geq 0$ such that \begin{align*} \mathfrak r_N\neq 0, \qquad \mathfrak r_{N+1}=0. \end{align*} Set \begin{align*} \mathfrak a:=\mathfrak r_N. \end{align*} Then $\mathfrak a\neq 0$ and \begin{align*} [\mathfrak a,\mathfrak a]=[\mathfrak r_N,\mathfrak r_N]=\mathfrak r_{N+1}=0, \end{align*} so $\mathfrak a$ is abelian. Each $\mathfrak r_m$ is an ideal of $\mathfrak g$. This is proved by induction: $\mathfrak r_0=\mathfrak r$ is an ideal, and if $\mathfrak r_m$ is an ideal, then for $z\in\mathfrak g$ and $[u,v]\in[\mathfrak r_m,\mathfrak r_m]$, the Jacobi identity gives \begin{align*} [z,[u,v]]=[[z,u],v]+[u,[z,v]]\in[\mathfrak r_m,\mathfrak r_m]. \end{align*} Thus $\mathfrak a=\mathfrak r_N$ is a nonzero abelian ideal of $\mathfrak g$.[/step]

[guided]Assume now that the Killing form is nondegenerate. We want to prove that $\mathfrak g$ is semisimple, so we must prove that its solvable radical is zero. Let \begin{align*} \mathfrak r:=\operatorname{rad}(\mathfrak g) \end{align*} denote the largest solvable ideal of $\mathfrak g$. Suppose, toward a contradiction, that $\mathfrak r\neq 0$. A nonzero solvable Lie algebra always has a last nonzero term in its derived series. Define the derived series by \begin{align*} \mathfrak r_0 &:= \mathfrak r,\\ \mathfrak r_{m+1} &:= [\mathfrak r_m,\mathfrak r_m]\quad \text{for } m\geq 0. \end{align*} Since $\mathfrak r$ is solvable, this series eventually reaches $0$. Since $\mathfrak r_0\neq 0$, there is an integer $N\geq 0$ such that \begin{align*} \mathfrak r_N\neq 0, \qquad \mathfrak r_{N+1}=0. \end{align*} Define \begin{align*} \mathfrak a:=\mathfrak r_N. \end{align*} Then $\mathfrak a$ is nonzero. Moreover, \begin{align*} [\mathfrak a,\mathfrak a] = [\mathfrak r_N,\mathfrak r_N] = \mathfrak r_{N+1} = 0, \end{align*} so $\mathfrak a$ is abelian. We also need $\mathfrak a$ to be an ideal of $\mathfrak g$, because the next step will use the action of arbitrary elements of $\mathfrak g$ on $\mathfrak a$. We prove that every $\mathfrak r_m$ is an ideal of $\mathfrak g$. The base case is $\mathfrak r_0=\mathfrak r$, which is an ideal by definition of the radical. Suppose $\mathfrak r_m$ is an ideal. If $z\in\mathfrak g$ and $u,v\in\mathfrak r_m$, then the Jacobi identity gives \begin{align*} [z,[u,v]]=[[z,u],v]+[u,[z,v]]. \end{align*} Because $\mathfrak r_m$ is an ideal, both $[z,u]$ and $[z,v]$ lie in $\mathfrak r_m$. Therefore both terms on the right lie in $[\mathfrak r_m,\mathfrak r_m]$, and hence \begin{align*} [z,[u,v]]\in[\mathfrak r_m,\mathfrak r_m]=\mathfrak r_{m+1}. \end{align*} Thus $\mathfrak r_{m+1}$ is an ideal of $\mathfrak g$. By induction, $\mathfrak a=\mathfrak r_N$ is a nonzero abelian ideal of $\mathfrak g$.[/guided]

[step:Show every element of a nonzero abelian ideal lies in the Killing kernel]Let $x\in\mathfrak a$ and $y\in\mathfrak g$. Consider the [linear map](/page/Linear%20Map) \begin{align*} T_{x,y}: \mathfrak g &\to \mathfrak g\\ z &\mapsto [x,[y,z]]. \end{align*} Then $T_{x,y}=\operatorname{ad}_x\circ\operatorname{ad}_y$. Since $\mathfrak a$ is an ideal, $\operatorname{ad}_y(\mathfrak a)\subseteq\mathfrak a$. Since $\mathfrak a$ is abelian and $x\in\mathfrak a$, $\operatorname{ad}_x$ vanishes on $\mathfrak a$. Hence $T_{x,y}$ restricts to the zero map on $\mathfrak a$. Also, since $\mathfrak a$ is an ideal and $x\in\mathfrak a$, we have $\operatorname{ad}_x(\mathfrak g)\subseteq\mathfrak a$. Therefore the induced map of $T_{x,y}$ on the quotient [vector space](/page/Vector%20Space) $\mathfrak g/\mathfrak a$ is zero. With respect to any basis adapted to the subspace $\mathfrak a\subseteq\mathfrak g$, the matrix of $T_{x,y}$ is block upper triangular with zero diagonal blocks. Hence \begin{align*} \operatorname{tr}(T_{x,y})=0. \end{align*} Therefore \begin{align*} \kappa_{\mathfrak g}(x,y) = \operatorname{tr}(\operatorname{ad}_x\circ\operatorname{ad}_y) = 0. \end{align*} Since $y\in\mathfrak g$ was arbitrary, $x\in\mathfrak k$. Thus $\mathfrak a\subseteq\mathfrak k$.[/step]

[guided]We now prove that the nonzero abelian ideal $\mathfrak a$ constructed above forces degeneracy of the Killing form. Fix $x\in\mathfrak a$ and $y\in\mathfrak g$. Define the linear map \begin{align*} T_{x,y}: \mathfrak g &\to \mathfrak g\\ z &\mapsto [x,[y,z]]. \end{align*} By definition of the adjoint representation, this is exactly \begin{align*} T_{x,y}=\operatorname{ad}_x\circ\operatorname{ad}_y. \end{align*} Thus \begin{align*} \kappa_{\mathfrak g}(x,y)=\operatorname{tr}(T_{x,y}). \end{align*} We compute this trace using the ideal $\mathfrak a$ as a filtration. First, because $\mathfrak a$ is an ideal, $\operatorname{ad}_y(\mathfrak a)\subseteq\mathfrak a$. Since $\mathfrak a$ is abelian and $x\in\mathfrak a$, the map $\operatorname{ad}_x$ vanishes on $\mathfrak a$. Therefore, if $z\in\mathfrak a$, then \begin{align*} T_{x,y}(z) = [x,[y,z]] = 0. \end{align*} So the restriction of $T_{x,y}$ to $\mathfrak a$ is the zero map. Second, because $\mathfrak a$ is an ideal and $x\in\mathfrak a$, we have $[x,w]\in\mathfrak a$ for every $w\in\mathfrak g$. Hence \begin{align*} T_{x,y}(\mathfrak g)\subseteq\mathfrak a. \end{align*} This means that the induced linear map on the quotient vector space $\mathfrak g/\mathfrak a$ is zero. Choose a basis of $\mathfrak a$ and extend it to a basis of $\mathfrak g$. With respect to this adapted basis, the matrix of $T_{x,y}$ is block upper triangular. Its diagonal block on $\mathfrak a$ is zero because $T_{x,y}|_{\mathfrak a}=0$, and its diagonal block on $\mathfrak g/\mathfrak a$ is zero because the induced quotient map is zero. Therefore its trace is zero: \begin{align*} \operatorname{tr}(T_{x,y})=0. \end{align*} Consequently, \begin{align*} \kappa_{\mathfrak g}(x,y) = \operatorname{tr}(\operatorname{ad}_x\circ\operatorname{ad}_y) = \operatorname{tr}(T_{x,y}) = 0. \end{align*} Since $y\in\mathfrak g$ was arbitrary, $x$ lies in the kernel $\mathfrak k$ of the Killing form. Because $x\in\mathfrak a$ was arbitrary, $\mathfrak a\subseteq\mathfrak k$.[/guided]

[step:Conclude that nondegeneracy is equivalent to semisimplicity] If $\kappa_{\mathfrak g}$ is nondegenerate, then $\mathfrak k=0$. The previous step gives $\mathfrak a\subseteq\mathfrak k$, while $\mathfrak a\neq 0$, a contradiction. Hence the assumption $\mathfrak r\neq 0$ is false, so $\operatorname{rad}(\mathfrak g)=0$. Therefore $\mathfrak g$ is semisimple. Combining this implication with the earlier implication that semisimplicity forces $\mathfrak k=0$, we obtain that $\mathfrak g$ is semisimple if and only if its Killing form is nondegenerate. [/step]