Theorem of Engel

LEMMA 5.15   Let $k$ be a commutative ring. Let $x \in M_n(k)$ be a nilpotent matrix. Then $\operatorname{ad}(x)$ is also nilpotent.

PROOF.. Assume $x^N=0$ . We decompose $\operatorname{ad}(x)$ into left and right multiplication. Namely,

$$\operatorname{ad}(x)=\lambda(x)-\rho(x)$$

Then $\lambda(x)$ and $\rho(x)$ commute with each other.

$$\operatorname{ad}(x)^{2N-1} =\sum_{j=0}^{2N-1}\lambda(x)^j(-\rho(x))^{2N-1-j} =\sum_{j=0}^{2N-1}\lambda(x^j)\rho((-x)^{2N-1-j})=0.$$

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THEOREM 5.16 (Engel)   Let $V$ be a finite dimensional vector space over a field $k$ . Let $L$ be a Lie subalgebra of $\mathfrak{gl}(V)$ such that each member of $L$ is a nilpotent matrix. Then:

1. If $\dim(L)\geq 1$ , then $L$ has an ideal of codimension $1$ .
2. If $\dim(V)\geq 1$ , then $L$ has a simultaneous 0 -eigen vector $v$ . (That is, $x.v=0 \qquad (\forall x\in L),\quad v\neq 0.$ )

PROOF.. If $\dim(L)=0$ , then there is nothing to do. We proceed by induction on $\dim(L)$ . Let $L_1$ be a maximal among

$$\{$$(Lie subalgebras of $L$ which is not equal to $L$)$$\}.$$

(The set above has a $\{0\}$ as a member, so it is not empty.) In view of the lemma above, we see

$$\forall x\in L_1 \exists N\in \mathbb{Z}_{>0} (\operatorname{ad}(x)^N(L)=0).$$

We note that an vector space $L/L_1$ admits adjoint actions by $L_1$ . Thus

$$\forall x\in L_1 \exists N\in \mathbb{Z}_{>0} (\operatorname{ad}(x)^N(L/L_1)=0).$$

That means, $\operatorname{ad}_{L/L_1}(L_1)\subset \mathfrak{gl}(L/L_1)$ also satisfy the assumption of the theorem. By the induction hypothesis, we see that conclusion (2) is applicable to this case. Namely, there exists an element $y_0\in L\setminus L_1$ such that

$$\operatorname{ad}(x)(y_0)(=[x,y_0])\in L_1 \qquad (\forall x \in L_1)$$

holds. Now a vector subset

$$L_2=k. y_0 +L_1 (\supsetneq L_1)$$

of $L$ is closed under Lie bracket and therefore it is a Lie subalgebra of $L$ . By the maximality of $L_1$ , $L_2$ should equal to $L$ . It is then also easy to verify that $L_1$ is an ideal of $L(=L_2)$ . This proves (1).

To prove (2), we note that $(L_1,V)$ satisfies the assumption of the theorem. So again by induction we see that $L_1$ has a simultaneous 0 -eigen vector. In other words,

$$V_1:=\bigcap_{x\in L_1} \{ v\in V; x.v=0\} \supsetneq \{0\}.$$

Let us then consider the action of $y_0$ .

$$v\in V_1 \implies x.(y_0.v)=y_0.(x.v)+[x,y_0].v =0 \implies y_0.v\in V_1$$

Thus $V_1$ admits an action of $y_0$ . Since $y_0$ is nilpotent on $V$ by the assumption, we see that $y_0$ has at least one 0 -eigen vector $v_0(\neq 0)$ in $V_1$ . Then $v_0$ surely is a simultaneous 0 -eigen vector of $L$ .

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THEOREM 5.17 (Engel)   Let $V$ be a finite dimensional vector space over a field $k$ . Let $L$ be a Lie subalgebra of $\mathfrak{gl}(V)$ such that each member of $L$ is a nilpotent matrix. Then there exists a basis $e_1,e_2,\dots, e_n$ of $V$ such that

$$L \subset \mathfrak{n}_n(k)=\left\{ \begin{pmatrix} 0 & * & * & ... ... \\ 0 & 0 & \dots & 0 & * \\ 0 & 0 & \dots & 0 & 0 \\ \end{pmatrix}\right\}$$

holds with respect to this basis. In particular, $L$ is nilpotent.

PROOF.. We apply the above theorem inductively to a vector space

$$L, L/(k. e_1), L/(k.e_1+k.e_2)\dots$$

and obtain the desired basis $\{e_i\}$ . Since the Lie algebra $\mathfrak{n}_n(k)$ is nilpotent, $L$ is also nilpotent.

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