Levi decomposition

DEFINITION 5.58   Let $L$ be a Lie algebra over a field $k$ . Let $R$ be the radical of $L$ . A Levi-subalgebra of $L$ is a subalgebra $L_1$ of $L$ such that $L$ is a direct sum of $L_1$ and $R$ as a vector space over $k$ .

We have the following obvious lemma.

LEMMA 5.59   Let $L$ be a Lie algebra over a field $k$ . Let $R$ be the radical of $L$ . Let $L_1$ be a Levi-subalgebra of $L$ . Then:

1. $L \cong L_1 \ltimes R$ .
2. $L_1\cong L/R$ .

In particular the isomorphism class of $L_1$ is unique.

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LEMMA 5.60   Let $n$ be a positive integer. Let $L$ be an $n$ -dimensional Lie algebra over a field $k$ of characteristic $p\in \operatorname{Ccs}(n^2)$ . Assume the radical $R$ of $L$ is abelian. Then:

1. $R=L^\perp$ . (We equip $L$ with the usual Killing form.)

2. $$\operatorname{Hom}_{k\operatorname{-linear}}(L,R)$$
   admits an action $\alpha$ of $L$ . Namely,
   $$(\alpha(x).\varphi)(y) =[x,\varphi(y)]-\varphi([x,y])\qquad (x\in L, y\in L).$$

3. For any $x\in L$ , we have $(\alpha(x).\varphi))\vert _R=0$ .
4. For any $x\in R$ , $\alpha(x)$ is nilpotent.

5. $$V_1=\{\varphi\in \operatorname{Hom}_{k\operatorname{-linear}}(L,R); \varphi\vert _R\in k\cdot\operatorname{id}_R \}$$
   is an $R$ -submodule of $\operatorname{Hom}_{k\operatorname{-linear}}(L,R)$ .
6. If $R$ is not equal to zero, then
   $$V_2=\{\varphi\in \operatorname{Hom}_{R\operatorname{-module}}(L,R); \varphi\vert _R\in k\cdot\operatorname{id}_R \}\neq 0.$$

7. There exists a Levi subalgebra $L_1$ of $L$ .

PROOF.. (1) Since $L^\perp$ has a trivial Killing form, it is solvable. (Cartan's criterion.) Thus, by the maximality of $R$ , we have

$$R\supset L^\perp.$$

On the other hand, let us take an arbitrary $x\in R$ , then for any $y,z\in L$ , we have

$$\operatorname{ad}(x)\operatorname{ad}(y)z=[x,[y,z]]\in [x,R] \subset R,$$

$$(\operatorname{ad}(x)\operatorname{ad}(y))^2 z\in [x,[y,R]]\subset [x,R]=0 \quad ($$Since $R$ is abelian.$$)$$

So $\kappa_L(x,y)=\operatorname{tr}_L(\operatorname{ad}(x)\operatorname{ad}(y))=0$ (as a trace of a nilpotent element.) Thus

$$R \subset L^\perp$$

also holds.

(2):follows from the general theory.

(3):follows easily from the definition of $\alpha$ .

(4):Let us take an arbitrary $x\in R$ . For any $y\in L$ , we have by using(2)

$$((\alpha(x)^2).\varphi)(y)=[x,(x.\varphi)(y)]-(x.\varphi)([x,y]) \in [R,R]-(x.\varphi)(R)=0.$$

Thus $\alpha(x)^2=0$ .

(5): follows clearly from (3).

(6): follows from (4),(5) and Engel's theorem. (We need to note that

$$V_2=\{ \varphi \in V_1; \forall x \in R (\alpha(x).\varphi=0 )\}$$

holds.) (7): The action of $R$ on $V_2$ is equal to zero. So $V_2$ admits an action by $L/R$ , which is semisimple. Now consider the following exact sequence of $L/R$ -modules.

$$0\to V_0\to V_2 \overset{\vert _R}{\to} k \to 0$$

where $V_0$ is the kernel of the restriction map. By a special case of Weyl's theorem on complete reducibility (Lemma 5.51), We see that the sequence splits. (Since we assumed $p\in \operatorname{Ccs}(n^2)$ ). This implies that there exists an element $\varphi_0\in V_2$ such that

$$\alpha(x).\varphi_0=0,\qquad \varphi_0\vert _R=\operatorname{id}_R.$$

Thus $\varphi_0$ gives a splitting of the injection $R\subset L$ . A Levi subalgebra of $L$ is obtained by putting

$$L_1=\{x-\varphi_0(x) ; x\in L\}.$$

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THEOREM 5.61 (Levi decomposition of a Lie algebra)   Let $n$ be a positive integer. Let $p \in \operatorname{Ccs}(n^2).$ Let $L$ be a $n$ -dimensional Lie algebra over a field of characteristic $p$ . Then $L$ has a Levi subalgebra $L_1$ . In other words, $L$ may be expressed as a semi direct product

$$L=L_1 \ltimes R$$

where $L_1$ is a semisimple (Levi) subalgebra of $L$ , and $R$ is a solvable (radical) ideal of $L$ .

PROOF.. If $R=0$ , then we only need to set $L_1=L$ . So let us assume $R\neq 0$ . Let us put

$$R_1=[R,R].$$

Then from the definition, we $R/R_1$ is an abelian Lie algebra. It is also easy to verify that $R_1$ is an ideal of $L$ . ($R_1$ is a characteristic ideal of $R$ ). We apply the preceding lemma for $R/R_1 \subset L/R_1$ to obtain a Levi subalgebra $M/R_1$ of $L/R_1$ . Then $M$ satisfies the following relations.

$$M + R=L,\quad M \cap R= R_1.$$

Since $R$ is solvable (and we have assumed $R\neq 0$ ), we see that $\dim(M)$ is strictly smaller than $\dim(L)$ . By induction $M$ have a Levi subalgebra $M_1$ . Then it is clear that $M_1$ is a Levi subalgebra of $L$ .

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