2.2 Engel's theorem

Engel's theorem

Let $g$ be a finite dimensional complex Lie algebra. Then $g$ is nilpotent $⟺$ $\mathrm{ad}X$ is nilpotent for each $X\in g$.

\begin{proof}i) Suppose $g$ is nilpotent. Then there is a $n$ such that $g^n=[g,[g,[\cdots ,[g,g]]\cdots ,]=0$. So for any $X_i\in g$, $\mathrm{ad}{X}_1\cdots \mathrm{ad}{X}_n=0$ and $(\mathrm{ad}X{)}^n=0$ for any $X\in g$.

ii) Conversely, suppose $\mathrm{ad}X$ is nilpotent for each $X\in g$. Claim: if $h⊊g$ is a nilpotent Lie subalgebra, then there is a $x\in g\backslash h$ such that $\tilde{h}=h\oplus \mathbb{C}X$ is a nilpotent Lie subalgebra. Obviously, for any non-zero $X_1\in g$, $\mathbb{C}{X}_1$ is a nilpotent Lie algebra. Then using the above claim, we can progressively get a nilpotent subalgebra sequence $h_1\subset h_2\subset \cdots \subset h_n$ until $h_n=g$. Now we prove the claim.

By the assumption of $\mathrm{ad}X$ nilpotent, the action $h\curvearrowright g$ (induced by adjoint representation) has only one generalized weight space, the one associated to the trivial weight. So the quotient action $h\curvearrowright g/h$ has one generalized weight space, the one associated to the trivial weight. Thus there exists non-zero $\tilde{X}\in g/h$ such that $\widetilde{\mathrm{ad}}(h)(\tilde{X})=0$, i.e. $[h,X]\subset h$ where $X+h=\tilde{X}$. So $\tilde{h}=h\oplus \mathbb{C}X$ is a subalgebra. Now we further argue that $\tilde{h}$ is nilpotent. Note that

\begin{aligned} &[\widetilde h,\widetilde h]=[h+\mathbb CX,h+\mathbb CX]\subseteq h\\ &[\widetilde h,h]=[h+\mathbb CX,h]\subseteq [h,h]+\mathbb C\cdot\mathrm{ad}X(h)\\ &[\widetilde h,[\widetilde h,[\widetilde h,\widetilde h]]]\subseteq [h,h]+\mathbb C\cdot(\mathrm{ad}X)^2(h)\\ &\cdots\\ &\text{so there is a $j$ s.t. $\widetilde h^j\subseteq h^2$ by $X$ nilpotent.} \end{aligned}\tag{*}$$ Claim: for each $i$, there is a $e(i)$ s.t. $\widetilde h^{e(i)}\subseteq h^i$. Show it by induction. By $(*)$ the case $k=1$ is true. Suppose $k=i$ is true, that is $\widetilde h^{e(i)}\subseteq h^i$. Then $$\begin{aligned} &\widetilde h^{e(i)+1}=[\widetilde h^{e(i)},\widetilde h]\subseteq[h^i,h+\mathbb CX]\subseteq h^{i+1}+\mathrm{ad}X(h^i)\\ &\widetilde h^{e(i)+2}=[\widetilde h^{e(i)+1},\widetilde h]\subseteq [h^{i+1}+\mathrm{ad}X(h^i),h+\mathbb CX]. \end{aligned}

By induction we can prove that $[X,h^i]\subseteq h^i$ for $i\in \mathbb{N}$. Then there is

$$\begin{array}{rl}& [h^{i+1}+\mathrm{ad}X(h^i),h+\mathbb{C}X]\\ \subseteq & [h^{i+1},h]+\mathbb{C}[h^{i+1},X]+[\mathrm{ad}X(h^i),h]+\mathbb{C}[\mathrm{ad}X(h^i),X]\\ \subseteq & h^{i+1}+\mathbb{C}[h^{i+1},X]+([X,[h^i,h]]+[h^i,[h,x]])+\mathbb{C}\cdot (\mathrm{ad}X{)}^2(h^i)\\ \subseteq & h^{i+1}+\mathbb{C}\cdot (\mathrm{ad}X{)}^2(h^i).\end{array}$$

Similarly, there is a big enough $j$ such that $(\mathrm{ad}X{)}^j=0$ and then

$${\tilde{h}}^{e(i)+j}\subseteq h^{i+1}+\mathbb{C}\cdot (\mathrm{ad}X{)}^j(h^i)=h^{i+1}.$$

By induction hypothesis, the claim is proved and so $\tilde{h}$ is nilpotent. \end{proof}