A <Solvable Lie algebra> is one whose derived series
$$
\mathfrak g^{(0)}=\mathfrak g,\qquad
\mathfrak g^{(r+1)}=[\mathfrak g^{(r)},\mathfrak g^{(r)}]
$$
eventually vanishes. By the <Lie theorem>, the adjoint operators of a solvable complex Lie algebra are simultaneously upper triangular. If $x\in[\mathfrak g,\mathfrak g]$, then $\operatorname{ad}_x$ is a sum of commutators of upper triangular matrices and is therefore strictly upper triangular. For every $y\in\mathfrak g$, the product $\operatorname{ad}_x\operatorname{ad}_y$ is strictly upper triangular, so
$$
\kappa(x,y)=0.
$$
Thus
$$
\boxed{[\mathfrak g,\mathfrak g]\subseteq\mathfrak g^\perp}.
$$

For a nonzero example, let $\mathfrak g$ have basis $h,e$ with $[h,e]=e$. It is solvable because $[\mathfrak g,\mathfrak g]=\mathbb Ce$ is abelian, but in the ordered basis $(h,e)$,
$$
\operatorname{ad}_h=
\begin{pmatrix}0&0\\0&1\end{pmatrix},
\qquad
\kappa(h,h)=1.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 102 / 1 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-102/1/c/solution