= Solution
A finite-dimensional complex <Lie algebra> is a <semisimple Lie algebra> when its <solvable radical> is zero, equivalently when it has no nonzero solvable ideals. Its <Killing form> is the <symmetric bilinear form>
$$
B_L(x,y)=\operatorname{tr}_L(\operatorname{ad}_x\operatorname{ad}_y).
$$
The cyclic <trace> identity and $\operatorname{ad}_{[z,x]}=[\operatorname{ad}_z,\operatorname{ad}_x]$ give
$$
B_L([z,x],y)+B_L(x,[z,y])=0.
$$
Thus the <Killing form> is an <invariant bilinear form on a Lie algebra>. In particular its radical $R=\{x:B_L(x,L)=0\}$ is an ideal.
We supply the <trace> argument needed for nondegeneracy rather than assuming the <Cartan criterion for semisimplicity>. The matrix form of the <Cartan solvability criterion> says: if $A\subseteq\operatorname{End}_{\mathbb C}(V)$ and $\operatorname{tr}(uv)=0$ for every $u\in[A,A]$, $v\in A$, then $A$ is solvable. To prove this direction, fix $u\in[A,A]$. Let $V=\bigoplus_\lambda V_\lambda$ be its <generalized eigenspace> decomposition, and define $s$ to act on $V_\lambda$ by the scalar $\overline\lambda$. On $\operatorname{Hom}(V_\mu,V_\lambda)$, the semisimple part of $\operatorname{ad}_u$ has <eigenvalue> $\lambda-\mu$, whereas $\operatorname{ad}_s$ acts by $\overline\lambda-\overline\mu=\overline{\lambda-\mu}$. <Polynomial interpolation> on the finitely many <eigenvalues>, with all derivatives through the sizes of the nilpotent blocks set to zero, therefore gives
$$
\operatorname{ad}_s=p(\operatorname{ad}_u),\qquad p(0)=0.
$$
Here the same difference always has the same conjugate, so the interpolation is consistent; the prescribed zero derivatives remove every nilpotent block. Since $\operatorname{ad}_u(A)\subseteq[A,A]$, it follows that $[s,A]\subseteq[A,A]$.
Write $u=\sum_j[a_j,b_j]$. By cyclicity and the <trace> hypothesis,
$$
\operatorname{tr}(us)=\sum_j\operatorname{tr}(a_j[b_j,s])=0.
$$
On the other hand, $\operatorname{tr}(us)=\sum_\lambda\dim(V_\lambda)|\lambda|^2$. Thus all <eigenvalues> of $u$ vanish, and every element of $[A,A]$ is nilpotent. The <Engel theorem> makes $[A,A]$ nilpotent as a <Lie algebra>, hence solvable; $A/[A,A]$ is abelian, so $A$ is solvable. This is the <Conjugate-spectrum proof of Cartan solvability>.
Apply this to the Killing radical $R$. For $x,y\in R$, the adjoint actions preserve $R$ and act as zero on $L/R$, because $R$ is an ideal. Computing traces in a basis adapted to $R$ gives
$$
B_R(x,y)=B_L(x,y)=0.
$$
The matrix <Lie algebra> $\operatorname{ad}_R(R)$ consequently satisfies the <trace> criterion and is solvable. Its kernel is $Z(R)$, an abelian ideal; a central extension of a solvable algebra is solvable. Thus $R$ is solvable. This proves the reusable assertion that the <Killing radical is a solvable ideal>. Semisimplicity forces $R=0$, and hence \b[$B_L$ is nondegenerate].
A <Cartan subalgebra> is a nilpotent subalgebra $H$ which is self-normalizing:
$$
N_L(H)=\{x\in L:[x,H]\subseteq H\}=H.
$$
For a complex semisimple algebra this is equivalently a maximal toral subalgebra. The nilpotent, self-normalizing definition permits a proof of the restricted nondegeneracy without first assuming the toral characterization.
Use the <generalized-weight decomposition for a nilpotent Lie algebra> for the adjoint action of $H$:
$$
L=\bigoplus_\lambda L^\lambda,\qquad
L^\lambda=\{v:(\operatorname{ad}_h-\lambda(h)I)^{\dim L}v=0\text{ for all }h\in H\}.
$$
For completeness, the stability underlying this decomposition follows directly from nilpotence of $H$. For fixed $h$, every $\operatorname{ad}_h$ on $H$ is nilpotent. If $T=\rho(h)$ and $S=\rho(y)$ in a finite-dimensional representation, then $(\operatorname{ad}_T)^rS=0$ for sufficiently large $r$. The identity
$$
(T-aI)^N S=\sum_{j=0}^N\binom Nj (\operatorname{ad}_T)^j(S)(T-aI)^{N-j}
$$
shows that $S$ preserves each <generalized eigenspace> of $T$. Starting with a basis of $H$, refine these primary decompositions successively; all summands remain $H$-invariant. Each resulting summand has only one <eigenvalue> for each basis element. The <Lie theorem> triangularizes the action on that summand, so those <eigenvalues> extend to a single linear character $\lambda$ on all of $H$. This proves the displayed decomposition and nilpotence of all shifted operators there.
Since $H$ is nilpotent, $H\subseteq L^0$. The space $L^0$ is a subalgebra: repeated use of the derivation rule for $\operatorname{ad}_h$ shows that the bracket of two generalized zero-<eigenvectors> is another such vector. If $L^0/H\ne0$, the adjoint action of $H$ on this quotient consists entirely of nilpotent maps. The <Engel theorem> supplies a nonzero coset $x+H$ with $[H,x]\subseteq H$, contradicting self-normalization. Therefore the <zero generalized weight space of a Cartan subalgebra> is exactly $H$.
Finally, $B_L(H,L^\lambda)=0$ when $\lambda\ne0$. Choose $h\in H$ with $\lambda(h)\ne0$ and put $T=\operatorname{ad}_h$. On $H$, a power $T^m$ vanishes. On $L^\lambda$, $T$ is invertible. For $u\in H$ and $v\in L^\lambda$, write $v=T^m w$; invariance gives
$$
B_L(u,v)=B_L(u,T^m w)=(-1)^mB_L(T^m u,w)=0.
$$
If $u\in H$ is orthogonal to $H$, it is now orthogonal to every summand of $L$, so nondegeneracy of $B_L$ implies $u=0$. \b[The restriction $B_L|_{H\times H}$ is nondegenerate.] This establishes the <nondegeneracy of the Killing form on a Cartan subalgebra> for every <Cartan subalgebra>, without requiring a chosen root basis.
Back to article page
