Past exam of the Lie algebras and representation theory of symmetric groups course of the University of Oxford / 2007 / A / 1 / b / Solution

We prove the common-zero-vector form of Engel's theorem by strong induction on the dimension of the acting Lie algebra of operators, allowing its action on a nonzero finite-dimensional space to have a kernel. The zero-dimensional case is immediate.

First, every proper Lie subalgebra $H<L$ has a strictly larger normalizer of a Lie subalgebra. The action of $H$ on $L/H$ by $h\cdot (x+H)=[h,x]+H$ consists of nilpotent operators by part (a). Its image has dimension at most $\dim H<\dim L$. By induction it has a nonzero common-zero vector $x+H$, giving $[H,x]\subseteq H$ with $x\notin H$. Hence $N_L(H)>H$.

Choose a maximal proper Lie subalgebra $H$. The preceding paragraph forces $N_L(H)=L$, so $H$ is an ideal of a Lie algebra. Moreover $L/H$ has dimension one: otherwise any nonzero one-dimensional subspace of $L/H$, which is a Lie subalgebra, would lift to an intermediate proper subalgebra. By induction the common kernelis nonzero. It is $L$-invariant, since $h(xv)=x(hv)+[h,x]v=0$ for $v\in W$, $h\in H$, $x\in L$. Take $x\notin H$. Its restriction to $W$ is nilpotent and so has a nonzero vector $v$ in its kernel. As $L=H+Kx$, this vector is killed by every element of $L$:This completes the induction. The underlying space is understood to be nonzero, as required for the asserted vector to exist.