Past exam of the Lie algebras course of the University of Oxford / 2011 / 1

Write the lower central series of a Lie algebra as ${\mathfrak{g}}^1=\mathfrak{g}$ and ${\mathfrak{g}}^{j+1}=[\mathfrak{g},{\mathfrak{g}}^j]$. Since the nilpotent Lie algebra is nonzero, there is a last nonzero term ${\mathfrak{g}}^c$, with ${\mathfrak{g}}^{c+1}=0$. Thus $[\mathfrak{g},{\mathfrak{g}}^c]=0$, soConsequently the centre is nonzero. This argument works over every field and includes an abelian algebra, for which $c=1$.

Fix $x$ with $x^N=0$ as a matrix. On the full matrix algebra define $L_x(y)=xy$ and $R_x(y)=yx$. These endomorphisms commute and each has $N$th power zero. The Adjoint representation of a Lie algebra is $\mathrm{ad}(x)=L_x-R_x$. The binomial theorem givesFor $m=2N-1$, each summand has either $j\ge N$ or $m-j\ge N$, and hence vanishes. Therefore $\mathrm{ad}(x)$ is nilpotent on the full matrix algebra and on its invariant subspace $\mathfrak{g}$.

Engel's theorem, in its adjoint form, states that a finite-dimensional Lie algebra is nilpotent if every $\mathrm{ad}(x)$ is a nilpotent endomorphism. Applying it to all $x\in \mathfrak{g}$ provesThis is distinct from assuming that an arbitrary representation of a nilpotent algebra must act by nilpotent matrices, which is false.

If $\mathfrak{g}$ is nilpotent, every Lie subalgebra $\mathfrak{a}$ is nilpotent, since ${\mathfrak{a}}^j\subseteq {\mathfrak{g}}^j$ in the respective lower central series of a Lie algebra. A two-dimensional nilpotent algebra has nonzero centre by part (i). If $0\ne z$ is central, complete it to a basis $(z,u)$; every bracket of these basis elements is zero. Thus $\mathfrak{a}$ is abelian.

Conversely, suppose $\mathfrak{g}$ is not nilpotent. By Engel's theorem, some $\mathrm{ad}(x)$ is not nilpotent. Since the field is algebraically closed, it has an eigenvalue $\lambda \ne 0$ with eigenvector $y\ne 0$. ThusThe vectors $x,y$ are independent: otherwise their bracket would be zero. Their span is therefore a two-dimensional Lie subalgebra and it is nonabelian. This contradicts the hypothesis. Hence

Suppose $\mathrm{ad}(x)$ has a nonzero eigenvalue $\lambda$ and corresponding vector $y$. Then $[x,y]=\lambda y$, soFor a diagonalizable endomorphism $T$, its eigenspace decomposition shows $\ker T^2=\ker T$. The displayed equations therefore contradict diagonalizability of $\mathrm{ad}(y)$.

Thus no $\mathrm{ad}(x)$ has a nonzero eigenvalue. Each is diagonalizable over the algebraically closed field, so each is the zero endomorphism. Hence $[x,z]=0$ for every $x,z\in \mathfrak{g}$, proving