Cartan's solvability criterion

Suppose first that $\mathfrak{h}$ is solvable. Choose bases of $\mathfrak{h}$ and $V$, and let $k_0\subseteq k$ be the subfield generated over $\mathbb{Q}$ by the finitely many structure constants and matrix entries. The resulting $k_0$-form ${\mathfrak{h}}_0$ is solvable because its derived series becomes that of $\mathfrak{h}$ after the faithful scalar extension $k_0\subseteq k$. The finitely generated field $k_0$ embeds in $\mathbb{C}$. Over $\mathbb{C}$, [L1] gives a basis in which ${\mathfrak{h}}_0$ is upper triangular and its derived algebra is strictly upper triangular. Thus $\mathrm{tr}(yx)=0$ for all $x\in [{\mathfrak{h}}_0,{\mathfrak{h}}_0]$ and $y\in {\mathfrak{h}}_0$. These finitely many bilinear identities hold over $k_0$ and hence after extension to $k$.

We prove the converse first under the stronger hypothesis $\mathrm{tr}(xy)=0$ for every $x,y\in \mathfrak{h}$. Choose bases and let $k_0$ be the characteristic-zero subfield generated by the finitely many matrix entries of a basis of $\mathfrak{h}$ and its structure constants. The strong trace hypothesis is determined by the finitely many pairs of basis vectors, so it persists after extending the resulting $k_0$-form to $\mathbb{C}$. Solvability over $\mathbb{C}$ descends because every derived term commutes with scalar extension. We may therefore carry out the strong converse over $\mathbb{C}$.

Over $\mathbb{C}$, fix $x\in [\mathfrak{h},\mathfrak{h}]$. By [L6], choose a Jordan basis on $V$ and write $x=x_s+x_n$, where $x_s$ is diagonal, $x_n$ is block-nilpotent, and they commute. The commuting left and right multiplication operators on $\mathrm{End}(V)$ show that the semisimple part of ${\mathrm{ad}}_x=L_x-R_x$ is ${\mathrm{ad}}_{x_s}$. Write the diagonal entries of $x_s$ as $a_1,\dots ,a_n$, and let $\overline{x_s}$ have diagonal entries $\overline{a_1},\dots ,\overline{a_n}$. On the generalized eigenspaces of ${\mathrm{ad}}_x$, ${\mathrm{ad}}_{\overline{x_s}}$ has eigenvalues $\overline{a_i-a_j}$. Finite Hermite interpolation on the finitely many differences $a_i-a_j$, taking zero to zero, expresses ${\mathrm{ad}}_{\overline{x_s}}$ as a polynomial without constant term in ${\mathrm{ad}}_x$. Since $\mathfrak{h}$ is stable under ${\mathrm{ad}}_x$, it follows that $[\overline{x_s},\mathfrak{h}]\subseteq \mathfrak{h}$.

Write $x$ as a finite sum of commutators. By [L5], $\mathrm{tr}(A[B,C])=\mathrm{tr}([A,B]C)$. Step 1.3 and the strong trace hypothesis therefore give $\mathrm{tr}(\overline{x_s}x)=0$. In the Jordan basis this trace is ${\sum }_i\overline{a_i}{a}_i$, so every $a_i=0$ and $x$ is nilpotent. If $x^m=0$ on $V$, then the commuting operators $L_x,R_x$ on $\mathrm{End}(V)$ satisfy $(L_x-R_x{)}^{2m-1}=0$, because every binomial term contains either $L_x^m$ or $R_x^m$. Restriction to the invariant derived algebra shows that each of its adjoint operators is nilpotent. Hence [L2] makes $[\mathfrak{h},\mathfrak{h}]$ nilpotent, and [L3] makes $\mathfrak{h}$ solvable.

Under the stated weaker hypothesis, apply steps 1.2–2.1 to the linear Lie algebra $[\mathfrak{h},\mathfrak{h}]$: the required strong trace vanishing holds for every pair in it. Its solvability implies solvability of $\mathfrak{h}$ by the derived-series definition. Together with step 1.1 this proves the linear criterion.

Apply the linear criterion to $\mathrm{ad}(\mathfrak{g})$. Its derived algebra is $\mathrm{ad}([\mathfrak{g},\mathfrak{g}])$, and [L7] says that the trace hypothesis is precisely the asserted Killing-form condition. Hence $\mathrm{ad}(\mathfrak{g})$ is solvable. The exact sequence $0\to Z(\mathfrak{g})\to \mathfrak{g}\to \mathrm{ad}(\mathfrak{g})\to 0$ has abelian kernel, so [L4] makes $\mathfrak{g}$ solvable. The reverse direction is step 1.1 for the adjoint representation. The zero algebra makes both conditions vacuous and is solvable.