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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,

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

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

$\displaystyle \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.
$

$ \qedsymbol$

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

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

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

$\displaystyle \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

$\displaystyle \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

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

holds. Now a vector subset

$\displaystyle 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,

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

Let us then consider the action of $ y_0$ .

$\displaystyle 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$ .

$ \qedsymbol$

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

$\displaystyle 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

$\displaystyle 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.

$ \qedsymbol$


next up previous
Next: Ideals of Up: generalities in finite dimensional Previous: The radicals of Lie
2007-12-19