[proofplan] We prove first that the Killing-orthogonal complement is stable under the adjoint action of $\mathfrak g$, hence is an ideal. The key computation uses the [invariance of the Killing form](/theorems/3808) and the fact that $\mathfrak a$ is an ideal. We then show that $\mathfrak a \cap \mathfrak a^\perp = 0$ by applying Cartan's criterion to this intersection, and the direct-sum decomposition follows from nondegeneracy and dimension counting. Finally, the commutator $[\mathfrak a,\mathfrak a^\perp]$ lies in both summands, so it must vanish. [/proofplan] [step:Verify the invariance identity for the Killing form] For $u,v,w \in \mathfrak g$, let \begin{align*} \operatorname{ad}: \mathfrak g &\to \mathfrak{gl}(\mathfrak g) \\ x &\mapsto \operatorname{ad}_x \end{align*} be the adjoint representation, where $\operatorname{ad}_x(y) := [x,y]$ for $y \in \mathfrak g$. The Killing form is \begin{align*} \kappa_{\mathfrak g}(x,y) := \operatorname{tr}(\operatorname{ad}_x \circ \operatorname{ad}_y). \end{align*} Using $\operatorname{ad}_{[u,v]} = \operatorname{ad}_u \circ \operatorname{ad}_v - \operatorname{ad}_v \circ \operatorname{ad}_u$ and cyclicity of trace on the finite-dimensional $F$-[vector space](/page/Vector%20Space) $\mathfrak g$, we obtain \begin{align*} \kappa_{\mathfrak g}([u,v],w) &= \operatorname{tr}\big((\operatorname{ad}_u \circ \operatorname{ad}_v - \operatorname{ad}_v \circ \operatorname{ad}_u)\circ \operatorname{ad}_w\big) \\ &= \operatorname{tr}(\operatorname{ad}_u \circ \operatorname{ad}_v \circ \operatorname{ad}_w) - \operatorname{tr}(\operatorname{ad}_v \circ \operatorname{ad}_u \circ \operatorname{ad}_w) \\ &= \operatorname{tr}(\operatorname{ad}_u \circ \operatorname{ad}_v \circ \operatorname{ad}_w) - \operatorname{tr}(\operatorname{ad}_u \circ \operatorname{ad}_w \circ \operatorname{ad}_v) \\ &= \operatorname{tr}\big(\operatorname{ad}_u \circ (\operatorname{ad}_v \circ \operatorname{ad}_w - \operatorname{ad}_w \circ \operatorname{ad}_v)\big) \\ &= \operatorname{tr}(\operatorname{ad}_u \circ \operatorname{ad}_{[v,w]}) \\ &= \kappa_{\mathfrak g}(u,[v,w]). \end{align*} Thus $\kappa_{\mathfrak g}$ is invariant: \begin{align*} \kappa_{\mathfrak g}([u,v],w)=\kappa_{\mathfrak g}(u,[v,w]). \end{align*} [/step] [step:Show that $\mathfrak a^\perp$ is stable under brackets with $\mathfrak g$] Let $x \in \mathfrak a^\perp$, let $g \in \mathfrak g$, and let $a \in \mathfrak a$. We prove that $[g,x] \in \mathfrak a^\perp$ by testing it against the arbitrary element $a$. Using invariance of $\kappa_{\mathfrak g}$ twice and symmetry of the Killing form, we compute \begin{align*} \kappa_{\mathfrak g}([g,x],a) &= \kappa_{\mathfrak g}(g,[x,a]) \\ &= \kappa_{\mathfrak g}([x,a],g) \\ &= \kappa_{\mathfrak g}(x,[a,g]). \end{align*} Since $\mathfrak a \trianglelefteq \mathfrak g$, we have $[a,g] \in \mathfrak a$. Since $x \in \mathfrak a^\perp$, this gives \begin{align*} \kappa_{\mathfrak g}(x,[a,g])=0. \end{align*} Therefore $\kappa_{\mathfrak g}([g,x],a)=0$ for every $a \in \mathfrak a$, so $[g,x] \in \mathfrak a^\perp$. Hence \begin{align*} [\mathfrak g,\mathfrak a^\perp] \subset \mathfrak a^\perp, \end{align*} and $\mathfrak a^\perp \trianglelefteq \mathfrak g$ is an ideal. [guided] We want to prove that $\mathfrak a^\perp$ is an ideal. By definition, this means that whenever $x \in \mathfrak a^\perp$ and $g \in \mathfrak g$, the bracket $[g,x]$ again lies in $\mathfrak a^\perp$. Since $\mathfrak a^\perp$ is defined by orthogonality to every element of $\mathfrak a$, the correct test is to pair $[g,x]$ with an arbitrary $a \in \mathfrak a$ under the Killing form. Let $x \in \mathfrak a^\perp$, $g \in \mathfrak g$, and $a \in \mathfrak a$. The invariance identity proved above gives \begin{align*} \kappa_{\mathfrak g}([g,x],a) &= \kappa_{\mathfrak g}(g,[x,a]). \end{align*} This expression does not yet use the fact that $x$ is orthogonal to $\mathfrak a$, because $g$ is not assumed to lie in $\mathfrak a^\perp$. We therefore move the bracket once more. By symmetry of the Killing form and then invariance, \begin{align*} \kappa_{\mathfrak g}(g,[x,a]) &= \kappa_{\mathfrak g}([x,a],g) \\ &= \kappa_{\mathfrak g}(x,[a,g]). \end{align*} Now the ideal hypothesis is used: since $\mathfrak a \trianglelefteq \mathfrak g$, the bracket $[a,g]$ belongs to $\mathfrak a$. Since $x \in \mathfrak a^\perp$, the definition of $\mathfrak a^\perp$ gives \begin{align*} \kappa_{\mathfrak g}(x,[a,g])=0. \end{align*} Thus \begin{align*} \kappa_{\mathfrak g}([g,x],a)=0 \end{align*} for every $a \in \mathfrak a$. This is exactly the statement that $[g,x] \in \mathfrak a^\perp$. Therefore $\mathfrak a^\perp$ is closed under brackets with all elements of $\mathfrak g$, so $\mathfrak a^\perp$ is an ideal of $\mathfrak g$. [/guided] [/step] [step:Prove that the two ideals have zero intersection] Let \begin{align*} \mathfrak b := \mathfrak a \cap \mathfrak a^\perp. \end{align*} Since both $\mathfrak a$ and $\mathfrak a^\perp$ are ideals of $\mathfrak g$, their intersection $\mathfrak b$ is an ideal of $\mathfrak g$. We claim that $\mathfrak b=0$. For $x,y \in \mathfrak b$, the adjoint maps $\operatorname{ad}_x$ and $\operatorname{ad}_y$ preserve $\mathfrak b$, because $\mathfrak b$ is an ideal. Let $\kappa_{\mathfrak b}: \mathfrak b \times \mathfrak b \to F$ denote the Killing form of $\mathfrak b$. Since $x,y \in \mathfrak b \subset \mathfrak a$ and also $y \in \mathfrak b \subset \mathfrak a^\perp$, we have \begin{align*} \kappa_{\mathfrak g}(x,y)=0. \end{align*} For an ideal $\mathfrak b \trianglelefteq \mathfrak g$, the restriction of $\kappa_{\mathfrak g}$ to $\mathfrak b \times \mathfrak b$ equals $\kappa_{\mathfrak b}$: indeed, for $x,y \in \mathfrak b$, the induced endomorphism of $\mathfrak g/\mathfrak b$ by $\operatorname{ad}_x \circ \operatorname{ad}_y$ is zero, because $\operatorname{ad}_y(\mathfrak g) \subset \mathfrak b$. Hence the trace on $\mathfrak g$ equals the trace on the invariant subspace $\mathfrak b$. Therefore \begin{align*} \kappa_{\mathfrak b}(x,y)=\kappa_{\mathfrak g}(x,y)=0 \end{align*} for every $x,y \in \mathfrak b$. Thus the Killing form of $\mathfrak b$ is identically zero. By Cartan's solvability criterion in characteristic $0$ (citing a result not yet in the wiki: [Cartan's criterion for solvability](/theorems/3811)), $\mathfrak b$ is solvable. Since $\mathfrak b$ is a solvable ideal of the semisimple Lie algebra $\mathfrak g$, semisimplicity forces $\mathfrak b=0$. Hence \begin{align*} \mathfrak a \cap \mathfrak a^\perp = 0. \end{align*} [/step] [step:Use nondegeneracy and dimension counting to obtain the direct sum] Because $\mathfrak g$ is semisimple over a field of characteristic $0$, its Killing form $\kappa_{\mathfrak g}$ is nondegenerate by Cartan's criterion for semisimplicity (citing a result not yet in the wiki: Cartan's criterion for semisimplicity via the Killing form). For a nondegenerate [bilinear form](/page/Bilinear%20Form) on the finite-dimensional vector space $\mathfrak g$, the orthogonal complement of the subspace $\mathfrak a$ satisfies \begin{align*} \dim_F \mathfrak a + \dim_F \mathfrak a^\perp = \dim_F \mathfrak g. \end{align*} Together with $\mathfrak a \cap \mathfrak a^\perp = 0$, this implies that the addition map \begin{align*} \mathfrak a \oplus \mathfrak a^\perp &\to \mathfrak g \\ (u,v) &\mapsto u+v \end{align*} is injective and has domain dimension equal to $\dim_F \mathfrak g$. Hence it is an isomorphism of $F$-vector spaces, and therefore \begin{align*} \mathfrak g = \mathfrak a \oplus \mathfrak a^\perp. \end{align*} Since both summands are ideals, this is a direct-sum decomposition by ideals. [/step] [step:Show that the two summands commute] Since $\mathfrak a \trianglelefteq \mathfrak g$, we have \begin{align*} [\mathfrak a,\mathfrak a^\perp] \subset \mathfrak a. \end{align*} Since $\mathfrak a^\perp \trianglelefteq \mathfrak g$, we also have \begin{align*} [\mathfrak a,\mathfrak a^\perp] \subset \mathfrak a^\perp. \end{align*} Therefore \begin{align*} [\mathfrak a,\mathfrak a^\perp] \subset \mathfrak a \cap \mathfrak a^\perp = 0. \end{align*} Thus \begin{align*} [\mathfrak a,\mathfrak a^\perp]=0. \end{align*} This proves all asserted conclusions. [/step]