= Solution
Work in finite dimension over $\mathbb C$; the characteristic-zero real version follows by complexification. The <Cartan solvability criterion> is
$$
\boxed{\mathfrak g\text{ is solvable}\quad\Longleftrightarrow\quad B(\mathfrak g,[\mathfrak g,\mathfrak g])=0.}
$$
Its semisimplicity consequence, also often called Cartan's criterion, is \b[$\mathfrak g$ is semisimple if and only if its <Killing form> is nondegenerate]. We prove both forms.
The auxiliary results used are as follows. The <Lie theorem> simultaneously triangularizes every finite-dimensional representation of a complex <solvable Lie algebra>. The <Engel theorem> says that a matrix <Lie algebra> consisting entirely of <nilpotent endomorphisms> has a common annihilated nonzero vector and a <basis> making all its matrices strictly upper triangular; in particular it is a <nilpotent Lie algebra>. Nilpotent means that the lower central series terminates, which implies that the derived series terminates as well. Finally, the <Additive Jordan decomposition> writes any complex endomorphism uniquely as $x=x_s+x_n$, where $x_s$ is diagonalizable, $x_n$ is nilpotent and they commute. On a <generalized eigenspace> $V_\lambda$, the two parts are $\lambda I$ and $x-\lambda I$, and both are polynomials in $x$. On endomorphisms, $\operatorname{ad}x_s$ is diagonalizable, $\operatorname{ad}x_n$ is nilpotent, and these are the Jordan parts of $\operatorname{ad}x$: the first assertion follows by splitting into maps between eigenspaces, and the second by expanding repeated <commutators> with the nilpotent $x_n$.
First prove the linear trace version: if $L\subseteq\mathfrak{gl}(V)$ and
$$
\operatorname{tr}(ab)=0\quad(a\in L,\ b\in[L,L]),
$$
then $L$ is solvable. Fix $x\in[L,L]$ and let $V=\bigoplus V_\lambda$ be its <generalized eigenspace> decomposition. Define $y$ to act on $V_\lambda$ by the scalar $\overline\lambda$. We do not assume that $y$, $x_s$ or $x_n$ belongs to $L$.
On $\operatorname{Hom}(V_\mu,V_\lambda)$, $\operatorname{ad}x$ has the form $(\lambda-\mu)I+N$ with $N$ nilpotent, whereas $\operatorname{ad}y$ is multiplication by $\overline{\lambda-\mu}$. Choose a polynomial $q$ with $q(d)=\bar d$ for every occurring <eigenvalue> difference $d$, and with all its derivatives of positive order up to the relevant nilpotency index zero at $d$. Such a polynomial exists by Hermite interpolation, or the Chinese remainder theorem for the pairwise coprime powers of $(t-d)$. At $d=0$ its value is zero, so $q(0)=0$. Expanding $q(dI+N)$ now gives
$$
\operatorname{ad}y=q(\operatorname{ad}x).
$$
Because $x\in L$ and $[L,L]$ is an ideal, every positive power of $\operatorname{ad}x$ sends $L$ into $[L,L]$. The zero constant term therefore gives $[y,L]\subseteq[L,L]$.
Express $x$ as a sum of <commutators> $\sum_j[a_j,b_j]$ with $a_j,b_j\in L$. Cyclicity of trace and the hypothesis imply
$$
\operatorname{tr}(xy)=\sum_j\operatorname{tr}([a_j,b_j]y)
=\sum_j\operatorname{tr}(a_j[b_j,y])=0.
$$
But on $V_\lambda$, $xy$ has trace $\dim(V_\lambda)|\lambda|^2$, since the nilpotent part has trace zero. Hence
$$
0=\operatorname{tr}(xy)=\sum_\lambda\dim(V_\lambda)|\lambda|^2,
$$
forcing every <eigenvalue> of $x$ to vanish. Thus every member of the <derived algebra> $[L,L]$ is a <nilpotent endomorphism>. By the <Engel theorem>, $[L,L]$ is nilpotent and therefore solvable; its derived series terminates, and adjoining the initial term $L$ shows that $L$ is solvable.
Conversely, if $L$ is solvable, the <Lie theorem> makes it upper triangular. <Commutators> of upper-triangular matrices have zero diagonal, and multiplying such a <commutator> by an upper-triangular matrix still has zero diagonal. Therefore the trace condition holds. Apply this equivalence to $L=\operatorname{ad}\mathfrak g$. Its trace pairing is the <Killing form>, and solvability of its image is equivalent to solvability of $\mathfrak g$: the kernel is its abelian centre, so a terminating derived series in the quotient terminates after at most one more step in $\mathfrak g$. This proves the solvability criterion.
For the semisimplicity criterion, let $R$ be the radical of the <Killing form>. Invariance makes $R$ a <Lie algebra ideal>. For $x,y\in R$, their adjoint actions on $\mathfrak g/R$ vanish, so computing trace in a <basis> adapted to $R$ gives $B_R(x,y)=B_{\mathfrak g}(x,y)=0$. The solvability criterion applied to $R$ shows it is a solvable ideal. Thus a <semisimple Lie algebra> has $R=0$ and a nondegenerate <Killing form>.
Conversely, any nonzero solvable ideal has a last nonzero term $A$ in its derived series; this is an abelian ideal of $\mathfrak g$. For $x\in A$, $y\in\mathfrak g$, the map $\operatorname{ad}x\operatorname{ad}y$ sends $\mathfrak g$ into $A$ and vanishes on $A$, so has trace zero. Hence $A\subseteq R$. A nondegenerate <Killing form> therefore rules out nonzero solvable ideals, exactly the definition of semisimplicity.
For classification, nondegeneracy of the <Killing form> provides the duality and root-space pairings used to extract a reduced crystallographic <root system> from a <semisimple Lie algebra>. Classification then reduces to the finite <Dynkin diagrams> and reconstruction of the <Lie algebra> from its <root> data.
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