Engel, the trace criterion, and Killing nondegeneracy

If $a^q=0$ on $V$, the commuting operators of left and right multiplication by $a$ on $\mathrm{End}(V)$ give $({\mathrm{ad}}_a{)}^m(T)={\sum }_{j=0}^m(-1{)}^j\left(\genfrac{}{}{0px}{}{m}{j}\right)a^{m-j}T{a}^j\mathrm{.}$ For $m\ge 2q-1$ every summand is zero. Hence ${\mathrm{ad}}_a$ is nilpotent on any invariant subspace or quotient, in particular on a quotient of subalgebras stable under its action.

For a finite-dimensional Lie algebra $\mathfrak{g}$, trace cyclicity proves symmetry of $B$. Jacobi and trace cyclicity also give $B([x,y],z)=\mathrm{tr}([{\mathrm{ad}}_x,{\mathrm{ad}}_y]{\mathrm{ad}}_z)=\mathrm{tr}({\mathrm{ad}}_x[{\mathrm{ad}}_y,{\mathrm{ad}}_z])=B(x,[y,z])\mathrm{.}$ Consequently its radical $J=\{x:B(x,\mathfrak{g})=0\}$ is an ideal: $B([a,x],y)=-B(x,[a,y])=0$ for $x\in J$.

We record a finite spectral construction. By F3 and F1, an operator $x$ on a nonzero $V$ has a direct-sum decomposition $V={⨁}_{\lambda }{V}_{\lambda }$ with $x{\mathrm{\mid }}_{V_{\lambda }}=\lambda \mathrm{id}+n_{\lambda }$, each $n_{\lambda }$ nilpotent. Define $b$ to act on $V_{\lambda }$ by the scalar $\bar{\lambda }$. For a map in $\mathrm{Hom}(V_{\mu },V_{\lambda })$, ${\mathrm{ad}}_x$ equals $(\lambda -\mu )\mathrm{id}$ plus the difference of commuting nilpotent left and right actions. Its nilpotent part has exponent at most $N=2\dim V$, by the same binomial expansion as in step 1.1. For every distinct difference $a=\lambda -\mu$, F2 supplies a polynomial $R(z)$ with $R(z)\equiv \bar{a}(\mathrm{m}\mathrm{o}\mathrm{d}(z-a{)}^N)$. These ideals are pairwise comaximal: distinct linear factors generate the unit ideal, and expanding a sufficiently high power of such a unit expression proves the same for their $N$th powers. Therefore $R({\mathrm{ad}}_x)$ acts on this Hom space by $\bar{\lambda }-\bar{\mu }$, exactly ${\mathrm{ad}}_b$. The difference zero is among the interpolation nodes, so $R(0)=0$. Thus ${\mathrm{ad}}_b=R({\mathrm{ad}}_x),R(0)=0.$ Also $\mathrm{tr}(bx)={\sum }_{\lambda }(\dim V_{\lambda })\mathrm{\mid }\lambda {\mathrm{\mid }}^2$, since each nilpotent block has trace zero.

Prove the common-zero-vector assertion by induction on $\dim L$, for all finite-dimensional nonzero representation spaces on which every operator is nilpotent. The zero algebra is immediate. Choose a proper subalgebra $H\subset L$ of maximal dimension. For $a\in H$, step 1.1 makes its adjoint action on $L\mathrm{/}H$ nilpotent. Its image is a Lie algebra of dimension at most $\dim H<\dim L$, so the induction hypothesis gives a nonzero coset $x+H$ annihilated by every $a\in H$. Hence $[H,x]\subseteq H$, and $H+\mathbb{C}x$ is a subalgebra strictly containing $H$. Maximality forces $L=H+\mathbb{C}x$ and makes $H$ a codimension-one ideal.

The induction hypothesis applied to $H$ on $V$ makes $K=\{v:Hv=0\}$ nonzero. It is $x$-invariant because for $a\in H$, $a(xv)=x(av)+[a,x]v=0$. The nilpotent restriction of $x$ to $K$ has a nonzero kernel: take the last nonzero vector in the finite power string of any nonzero vector of $K$. This vector is killed by both $x$ and $H$, hence by $L$, completing the induction. Apply the same common-vector result to successive quotients of $V$ to obtain a finite invariant flag with zero action on each one-dimensional quotient. Lifting a basis of that flag gives strict upper triangularity. A product of $\dim V$ strictly upper triangular operators is zero; expanding iterated brackets into products shows this algebra, and every subalgebra of it, is solvable (indeed nilpotent).

Now suppose the trace-zero hypothesis holds and fix $x\in [L,L]$. Use the operator $b$ and polynomial $R$ from step 2.1. Write $x={\sum }_j[y_j,z_j]$ with $y_j,z_j\in L$, a finite sum by the definition of the derived subspace. Cyclicity of finite matrix trace gives $\mathrm{tr}(bx)={\sum }_j\mathrm{tr}([b,y_j]z_j)={\sum }_j\mathrm{tr}(R({\mathrm{ad}}_x)(y_j)z_j)=0.$ Indeed $[L,L]$ is an ideal by Jacobi, and $R(0)=0$ implies $R({\mathrm{ad}}_x)(y_j)\in [L,L]$, so each last trace vanishes by the hypothesis. Step 2.1 now gives a sum of nonnegative real numbers ${\sum }_{\lambda }(\dim V_{\lambda })\mathrm{\mid }\lambda {\mathrm{\mid }}^2=0$. Each $\lambda$ is zero, so $x$ is nilpotent on $V$.

Thus every element of $[L,L]$ is a nilpotent operator. Step 3.1 makes this derived algebra solvable. If its derived series vanishes after $m$ steps, that of $L$ vanishes after $m+1$, proving the trace criterion. If $V=0$, then $L=0$ and the same conclusion is immediate without spectral decomposition.

Suppose $\mathfrak{g}$ is semisimple. For $x,y\in J$, the adjoint actions preserve $J$ and act by zero on $\mathfrak{g}\mathrm{/}J$. Computing trace in a basis extending one of $J$ gives $B_J(x,y)=B_{\mathfrak{g}}(x,y)=0$. Thus $L={\mathrm{ad}}_J(J)\subseteq \mathrm{End}(J)$ satisfies step 4.1's trace hypothesis and is solvable. Its kernel in $J$ is the center, which is abelian; explicitly, if $D^{m}L=0$ then $D^{m}J$ lies in that center and $D^{m+1}J=0$. Hence $J$ is a solvable ideal of $\mathfrak{g}$ and is zero by semisimplicity. This proves nondegeneracy. For $\mathfrak{g}=0$ the unique form has zero radical. Every induction, basis, polynomial interpolation and eigenvalue factorization above is finite; no algebraic closure or arbitrary-index selection is taken, so no AC is used.