[proofplan] We identify the radical of the restricted form as $\mathfrak r=\mathfrak a \cap \mathfrak a^\perp$, where orthogonality is taken with respect to $\kappa_{\mathfrak g}$. [Invariance of the Killing form](/theorems/3808) shows that both $\mathfrak a^\perp$ and $\mathfrak r$ are ideals of $\mathfrak g$. We then prove that $\kappa_{\mathfrak g}(x,y)=0$ for every $x \in [\mathfrak r,\mathfrak r]$ and $y \in \mathfrak r$, so Cartan's solvability criterion applied to the adjoint representation of $\mathfrak r$ on $\mathfrak g$ implies that $\mathfrak r$ is solvable. Since $\mathfrak g$ is semisimple, it has no nonzero solvable ideals, hence $\mathfrak r=0$, which is exactly the nondegeneracy of the restriction. [/proofplan]

[step:Identify the radical of the restricted Killing form]Define the orthogonal complement of $\mathfrak a$ in $\mathfrak g$ with respect to $\kappa_{\mathfrak g}$ by \begin{align*} \mathfrak a^\perp := \{x \in \mathfrak g : \kappa_{\mathfrak g}(x,y)=0 \text{ for every } y \in \mathfrak a\}. \end{align*} Define \begin{align*} \mathfrak r := \mathfrak a \cap \mathfrak a^\perp. \end{align*} Then $\mathfrak r$ is precisely the radical of the restricted [bilinear form](/page/Bilinear%20Form) $\kappa_{\mathfrak g}|_{\mathfrak a \times \mathfrak a}$: an element $x \in \mathfrak a$ lies in the radical iff $\kappa_{\mathfrak g}(x,y)=0$ for every $y \in \mathfrak a$, iff $x \in \mathfrak a^\perp$, iff $x \in \mathfrak r$.[/step]

[guided]The restricted form is nondegenerate exactly when its radical is zero. We therefore name that radical explicitly. Define the orthogonal complement of $\mathfrak a$ in $\mathfrak g$ with respect to the ambient Killing form by \begin{align*} \mathfrak a^\perp := \{x \in \mathfrak g : \kappa_{\mathfrak g}(x,y)=0 \text{ for every } y \in \mathfrak a\}. \end{align*} Then define \begin{align*} \mathfrak r := \mathfrak a \cap \mathfrak a^\perp. \end{align*} If $x \in \mathfrak a$, then $x$ lies in the radical of $\kappa_{\mathfrak g}|_{\mathfrak a \times \mathfrak a}$ precisely when \begin{align*} \kappa_{\mathfrak g}(x,y)=0 \quad \text{for every } y \in \mathfrak a. \end{align*} This condition is exactly $x \in \mathfrak a^\perp$. Since we already require $x \in \mathfrak a$, the radical is exactly $\mathfrak a \cap \mathfrak a^\perp=\mathfrak r$. Thus proving nondegeneracy is equivalent to proving $\mathfrak r=0$.[/guided]

[step:Use invariance to show that the orthogonal complement is an ideal]We show that $\mathfrak a^\perp \trianglelefteq \mathfrak g$. Let $z \in \mathfrak a^\perp$, let $y \in \mathfrak g$, and let $t \in \mathfrak a$. Since $\mathfrak a$ is an ideal, $[y,t] \in \mathfrak a$. By invariance and symmetry of the Killing form, \begin{align*} \kappa_{\mathfrak g}([y,z],t) &= -\kappa_{\mathfrak g}(z,[y,t]) =0. \end{align*} Since this holds for every $t \in \mathfrak a$, we have $[y,z] \in \mathfrak a^\perp$. Therefore $\mathfrak a^\perp$ is an ideal of $\mathfrak g$. Because $\mathfrak a$ is also an ideal, their intersection $\mathfrak r=\mathfrak a \cap \mathfrak a^\perp$ is an ideal of $\mathfrak g$.[/step]

[guided]We need $\mathfrak r$ to be an ideal because semisimplicity rules out nonzero solvable ideals. Since $\mathfrak r$ is an intersection, it is enough to show that $\mathfrak a^\perp$ is an ideal. Let $z \in \mathfrak a^\perp$, let $y \in \mathfrak g$, and let $t \in \mathfrak a$. To prove $[y,z] \in \mathfrak a^\perp$, we must prove that $\kappa_{\mathfrak g}([y,z],t)=0$ for every $t \in \mathfrak a$. The invariance of the Killing form says that bracket operations may be moved from one argument to the other with the corresponding sign convention. Using invariance and symmetry, we get \begin{align*} \kappa_{\mathfrak g}([y,z],t) &= -\kappa_{\mathfrak g}(z,[y,t]). \end{align*} Now $\mathfrak a$ is an ideal, so $[y,t] \in \mathfrak a$. Since $z \in \mathfrak a^\perp$, it is orthogonal to every element of $\mathfrak a$. Therefore \begin{align*} \kappa_{\mathfrak g}([y,z],t) &= -\kappa_{\mathfrak g}(z,[y,t]) =0. \end{align*} This holds for all $t \in \mathfrak a$, so $[y,z] \in \mathfrak a^\perp$. Hence $\mathfrak a^\perp \trianglelefteq \mathfrak g$. Since intersections of ideals are ideals, $\mathfrak r=\mathfrak a \cap \mathfrak a^\perp$ is an ideal of $\mathfrak g$.[/guided]

[step:Show that the Killing form vanishes on $[\mathfrak r,\mathfrak r]\times \mathfrak r$] Let $x \in [\mathfrak r,\mathfrak r]$ and $y \in \mathfrak r$. It is enough to check generators of $[\mathfrak r,\mathfrak r]$, so write $x=[u,v]$ with $u,v \in \mathfrak r$. Since $\mathfrak r$ is an ideal of $\mathfrak g$, $[v,y] \in \mathfrak r$. Since $\mathfrak r \subseteq \mathfrak a$ and $u \in \mathfrak r \subseteq \mathfrak a^\perp$, invariance gives \begin{align*} \kappa_{\mathfrak g}(x,y) &= \kappa_{\mathfrak g}([u,v],y) \\ &= \kappa_{\mathfrak g}(u,[v,y]) \\ &=0. \end{align*} By linearity, $\kappa_{\mathfrak g}(x,y)=0$ for every $x \in [\mathfrak r,\mathfrak r]$ and every $y \in \mathfrak r$.[/step]

[guided]The goal is to prepare for Cartan's solvability criterion, whose trace condition involves elements of the derived algebra $[\mathfrak r,\mathfrak r]$. So take $x \in [\mathfrak r,\mathfrak r]$ and $y \in \mathfrak r$. Because $[\mathfrak r,\mathfrak r]$ is linearly spanned by brackets $[u,v]$ with $u,v \in \mathfrak r$, it is enough by bilinearity of $\kappa_{\mathfrak g}$ to prove the vanishing when $x=[u,v]$. Let $u,v \in \mathfrak r$. Since $\mathfrak r$ is an ideal of $\mathfrak g$ and $y \in \mathfrak r \subseteq \mathfrak g$, we have \begin{align*} [v,y] \in \mathfrak r. \end{align*} In particular $[v,y] \in \mathfrak a$, because $\mathfrak r \subseteq \mathfrak a$. Also $u \in \mathfrak a^\perp$, because $\mathfrak r \subseteq \mathfrak a^\perp$. The invariance of the Killing form gives \begin{align*} \kappa_{\mathfrak g}([u,v],y) &= \kappa_{\mathfrak g}(u,[v,y]). \end{align*} The right-hand side is zero because $u$ is orthogonal to every element of $\mathfrak a$ and $[v,y] \in \mathfrak a$. Therefore \begin{align*} \kappa_{\mathfrak g}([u,v],y)=0. \end{align*} Linearity extends the conclusion from bracket generators to all $x \in [\mathfrak r,\mathfrak r]$.[/guided]

[step:Apply Cartan's solvability criterion to the adjoint representation of $\mathfrak r$]Define the adjoint representation of $\mathfrak g$ by \begin{align*} \operatorname{ad}_{\mathfrak g}: \mathfrak g &\to \mathfrak{gl}(\mathfrak g) \\ x &\mapsto \bigl(y \mapsto [x,y]\bigr). \end{align*} For $x,y \in \mathfrak g$, the Killing form is \begin{align*} \kappa_{\mathfrak g}(x,y) = \operatorname{tr}_{\mathfrak g}\bigl(\operatorname{ad}_{\mathfrak g}(x)\circ \operatorname{ad}_{\mathfrak g}(y)\bigr). \end{align*} The previous step gives this trace equal to $0$ for every $x \in [\mathfrak r,\mathfrak r]$ and $y \in \mathfrak r$. Because $\mathfrak g$ is semisimple, its center is zero: any central element spans an abelian solvable ideal. Hence $\operatorname{ad}_{\mathfrak g}|_{\mathfrak r}$ is injective. Applying Cartan's solvability criterion for linear Lie algebras (citing a result not yet in the wiki: Cartan's solvability criterion) to the Lie subalgebra $\operatorname{ad}_{\mathfrak g}(\mathfrak r) \subseteq \mathfrak{gl}(\mathfrak g)$ shows that $\operatorname{ad}_{\mathfrak g}(\mathfrak r)$ is solvable. Since $\operatorname{ad}_{\mathfrak g}|_{\mathfrak r}$ is an isomorphism from $\mathfrak r$ onto its image, $\mathfrak r$ is solvable.[/step]

[guided]We now convert the orthogonality computation into solvability. Define the adjoint representation \begin{align*} \operatorname{ad}_{\mathfrak g}: \mathfrak g &\to \mathfrak{gl}(\mathfrak g) \\ x &\mapsto \bigl(y \mapsto [x,y]\bigr). \end{align*} By definition of the Killing form, \begin{align*} \kappa_{\mathfrak g}(x,y) = \operatorname{tr}_{\mathfrak g}\bigl(\operatorname{ad}_{\mathfrak g}(x)\circ \operatorname{ad}_{\mathfrak g}(y)\bigr) \end{align*} for all $x,y \in \mathfrak g$. From the previous step, this trace is zero whenever $x \in [\mathfrak r,\mathfrak r]$ and $y \in \mathfrak r$. Cartan's solvability criterion for linear Lie algebras says that a finite-dimensional Lie subalgebra $\mathfrak l \subseteq \mathfrak{gl}(V)$ over a field of characteristic $0$ is solvable if \begin{align*} \operatorname{tr}_{V}(AB)=0 \end{align*} for every $A \in [\mathfrak l,\mathfrak l]$ and every $B \in \mathfrak l$ (citing a result not yet in the wiki: Cartan's solvability criterion). We apply this criterion to \begin{align*} \mathfrak l := \operatorname{ad}_{\mathfrak g}(\mathfrak r) \subseteq \mathfrak{gl}(\mathfrak g). \end{align*} The commutator algebra of this image is \begin{align*} [\mathfrak l,\mathfrak l] = \operatorname{ad}_{\mathfrak g}([\mathfrak r,\mathfrak r]), \end{align*} because $\operatorname{ad}_{\mathfrak g}$ is a Lie algebra homomorphism. Thus the trace condition supplied by the previous step is exactly the trace condition required by Cartan's criterion. Therefore $\operatorname{ad}_{\mathfrak g}(\mathfrak r)$ is solvable. Finally we pass solvability back from the image to $\mathfrak r$. Since $\mathfrak g$ is semisimple, its center is zero: if $z \in Z(\mathfrak g)$, then $Fz$ is an abelian ideal of $\mathfrak g$, hence a solvable ideal, so semisimplicity forces $z=0$. Therefore the kernel of $\operatorname{ad}_{\mathfrak g}$ is $Z(\mathfrak g)=0$, and the restriction $\operatorname{ad}_{\mathfrak g}|_{\mathfrak r}$ is injective. It is therefore an isomorphism from $\mathfrak r$ onto $\operatorname{ad}_{\mathfrak g}(\mathfrak r)$. Since solvability is preserved under Lie algebra isomorphism, $\mathfrak r$ is solvable.[/guided]

[step:Use semisimplicity to eliminate the radical]We have shown that $\mathfrak r$ is a solvable ideal of $\mathfrak g$. Since $\mathfrak g$ is semisimple, its radical is zero, equivalently $\mathfrak g$ has no nonzero solvable ideals. Hence $\mathfrak r=0$. By the first step, $\mathfrak r$ is the radical of $\kappa_{\mathfrak g}|_{\mathfrak a \times \mathfrak a}$. Therefore the restriction $\kappa_{\mathfrak g}|_{\mathfrak a \times \mathfrak a}$ is nondegenerate.[/step]

[guided]The construction of $\mathfrak r$ was designed so that there is only one possible final obstruction: $\mathfrak r$ could be a nonzero radical of the restricted form. But the previous step shows that $\mathfrak r$ is solvable, and the second step shows that $\mathfrak r$ is an ideal of $\mathfrak g$. By semisimplicity, the radical of $\mathfrak g$, meaning its largest solvable ideal, is zero. Equivalently, $\mathfrak g$ has no nonzero solvable ideals. Since $\mathfrak r$ is a solvable ideal of $\mathfrak g$, it follows that \begin{align*} \mathfrak r=0. \end{align*} The first step identified $\mathfrak r$ with the radical of the restricted bilinear form $\kappa_{\mathfrak g}|_{\mathfrak a \times \mathfrak a}$. Thus the restricted form has zero radical, which is precisely nondegeneracy.[/guided]