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

$$\begin{array}{rl}& [\tilde{h},\tilde{h}]=[h+\mathbb{C}X,h+\mathbb{C}X]\subseteq h\\ & [\tilde{h},h]=[h+\mathbb{C}X,h]\subseteq [h,h]+\mathbb{C}\cdot \mathrm{ad}X(h)\\ & [\tilde{h},[\tilde{h},[\tilde{h},\tilde{h}]]]\subseteq [h,h]+\mathbb{C}\cdot (\mathrm{ad}X{)}^2(h)\\ & \cdots \\ & \text{so there is a j s.t. hj⊆h2 by X nilpotent.}\end{array}\text{\hspace{1em}(*)}$$

Claim: for each $i$, there is a $e(i)$ s.t. ${\tilde{h}}^{e(i)}\subseteq h^i$. Show it by induction. By $(\ast )$ the case $k=1$ is true. Suppose $k=i$ is true, that is ${\tilde{h}}^{e(i)}\subseteq h^i$. Then

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

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}