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

On the space of endomorphisms of $V$, left and right multiplication by $x$ commute, and $\mathrm{ad}x$ is their difference. The binomial theorem therefore givesFor $m=2n-1$, every term has either $j\ge n$ or $m-j\ge n$, so every term vanishes when $x^n=0$. Restriction to the invariant Lie subalgebra $L$ provesThe argument works over every field; no division by a binomial coefficient is used.

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.

Start with the given basis of $U$, and follow it by arbitrary lifts of the given basis of $V/U$. Since $U$ is an invariant linear subspace, every acting endomorphism has block formwhere $A$ and $B$ are its matrices on $U$ and $V/U$. Both are strictly upper triangular in the chosen bases. Therefore the whole matrix has zero entries on and below its diagonal. The same lifted basis works simultaneously for all elements of $L$.

Use induction on $\dim V$. By part (b), there is a one-dimensional invariant linear subspace $U=Kv$ on which $L$ acts as zero. The induced operators on $V/U$ remain nilpotent, so induction gives a simultaneous strictly upper triangular basis there. Part (c) lifts it together with $v$ to such a basis of $V$.

Thus all elements strictly lower one common complete flag. A nested Lie bracket of sufficiently many flag-lowering operators lowers it past zero. Consequently the lower central series of $L$ terminates, proving that the Lie algebra is nilpotent. In fact its nilpotency class is at most $n-1$ for $n>1$; the zero algebra covers the one-dimensional case.