= Solution
A finite-dimensional <Lie algebra> over the <complex numbers> $L$ is a <semisimple Lie algebra> when its <solvable radical> is zero, equivalently when it has no nonzero solvable ideals. Its <Killing form> is
$$
B_L(x,y)=\operatorname{tr}_L(\operatorname{ad}x\operatorname{ad}y).
$$
The <cyclic property of the trace> makes this <bilinear form> symmetric and gives its <invariance of a bilinear form on a Lie algebra>:
$$
B_L([x,y],z)=B_L(x,[y,z]).
$$
It follows that $R=\{x:B_L(x,L)=0\}$ is an <ideal of a Lie algebra>. For $x\in R$, $\operatorname{ad}x$ induces the zero map on $L/R$. Therefore, for $x,y\in R$, the <matrix trace> splits over the invariant subspace $R$ and the quotient to give $B_R(x,y)=B_L(x,y)=0$. In particular $B_R([R,R],R)=0$. The <Cartan solvability criterion> implies that $R$ is solvable. Since $L$ is semisimple, $R=0$: \b[the Killing form is nondegenerate].
For completeness, the trace step in the <Cartan solvability criterion> is precisely the mechanism of the previous solution. For a complex matrix <Lie algebra> $\mathfrak a$ with $\operatorname{tr}(xy)=0$ for $x\in[\mathfrak a,\mathfrak a]$, $y\in\mathfrak a$, set $U=[\mathfrak a,\mathfrak a]$ and $W=\mathfrak a$. If $\beta\in M$ and $x=[a,b]$, then $\operatorname{tr}(x\beta)=\operatorname{tr}(a[b,\beta])=0$, since $[b,\beta]\in U$. Linearity and the <trace orthogonality nilpotence lemma> show that every member of $U$ is nilpotent. The <Engel theorem> makes $U$ nilpotent and hence $\mathfrak a$ solvable. Apply this to $\mathfrak a=\operatorname{ad}R$; the kernel of this <Adjoint representation of a Lie algebra> is the abelian center of $R$, so $R$ is solvable as claimed.
For an arbitrary complex <Lie algebra>, a <Cartan subalgebra> $H$ means a <nilpotent Lie algebra> that is self-normalizing: $N_L(H)=\{x:[x,H]\subseteq H\}=H$. This definition does not assume that $H$ is abelian. We prove that it is abelian when $L$ is semisimple.
Use the <generalized-weight decomposition for a nilpotent Lie algebra> for the action of $H$ on $L$. Its zero generalized <weight space> is
$$
L^0=\{x:(\operatorname{ad}h)^{\dim L}x=0\text{ for every }h\in H\}.
$$
We have $H\subseteq L^0$, since $H$ is nilpotent. If $L^0/H\ne0$, the <Engel theorem> gives a nonzero coset annihilated by every $h\in H$. Its representative $x$ satisfies $[H,x]\subseteq H$, contradicting $N_L(H)=H$. Thus $L^0=H$.
For a nonzero generalized <weight> $\lambda$, choose $h_0\in H$ with $\lambda(h_0)\ne0$. The operator $\operatorname{ad}h_0$ is invertible on $L^\lambda$ and nilpotent on $H$. For $h\in H$, $z\in L^\lambda$, write $z=(\operatorname{ad}h_0)^m w$ with $m$ large enough that $(\operatorname{ad}h_0)^m h=0$. Invariance of the <Killing form> gives
$$
B_L(h,z)=(-1)^mB_L((\operatorname{ad}h_0)^m h,w)=0.
$$
On the other hand, $H$ is solvable, so the <Lie theorem> triangularizes its action on $L$. For $a,b,h\in H$, the matrix $\operatorname{ad}[a,b]$ is strictly upper triangular, while $\operatorname{ad}h$ is upper triangular. Thus $B_L([H,H],H)=0$. Together with $L=H\oplus\bigoplus_{\lambda\ne0}L^\lambda$, this yields $B_L([H,H],L)=0$. Nondegeneracy gives $[H,H]=0$.
Finally, if $x$ commutes with $H$, it normalizes $H$, hence lies in $H$. Any abelian subalgebra containing $H$ consists of such elements. Thus \b[$H$ is a maximal abelian subalgebra], indeed $C_L(H)=H$.
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