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.

$\qedsymbol$

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

1. The radical $R$ of $L$ is abelian.
2. The center of $L$ is trivial.
3. $p>n^2$ or $p=0$ .

Then:

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

2. $$V_1=\{\varphi\in \operatorname{Hom}_{k\operatorname{-linear}}(L,R); \varphi\vert _R\in k\cdot\operatorname{id}_R \}$$
   is an $L$ -submodule of $V$ .

3. $$V_0=\{\varphi\in \operatorname{Hom}_{k\operatorname{-linear}}(L,R); \varphi\vert _R=0 \}$$
   is an $L$ -submodule of $V_1$ .

4. $$U=\operatorname{ad}_L(R)$$
   is an $L$ -submodule of $V_0$ .
5. $R.V_1 \subset U$ .
6. There exists $\chi\in \operatorname{Hom}_{k\operatorname{-linear}}(L,R)$ such that
   $$\chi\vert _R=\operatorname{id}_R, \qquad L.\chi \subset U$$
   holds.
7. For any $x\in L$ , there exists a unique element $r_x\in R$ such that
   $$x.\chi=\operatorname{ad}r_x$$
   holds.

8. $$h:L\ni x \mapsto -r_x \in R$$
   is a $k$ -linear projection.
9. $L_1=\operatorname{Ker}(h)$ is a Levi subalgebra of $L$ .

PROOF.. (1):follows from the general theory.

(2),(3),(4):follows easily from the definition of $\alpha$ .

(5): For any $r\in R$ and for any $\varphi\in V_1$ , we have

$$r.\phi=(z\mapsto [r,[\phi(z)]-\phi([r,z]))=(z\mapsto 0-c_\phi [r,z]) =\operatorname{ad}_L (-c_\phi r) \in U.$$

(6): $V_1/U$ is an $L/R$ -module which has $V_0/U$ as a submodule of codimension $1$ . Thus by using Lemma 5.51 (Weyl's theorem on irreducibility (codimension $1$ case)), we see that there exists a 1-dimensional $L/R$ -submodule $X/U$ of $V_1/U$ (where $X$ is a $L$ submodule of $L$ ) such that

$$V_1/U=V_0/U\oplus X/U$$

holds. Since $X/U$ is $1$ dimensional, the action of the semisimple Lie algebra $L/R$ on $X/U$ is trivial. Thus we see that there exists an element $\chi\in \operatorname{Hom}_{k\operatorname{-linear}}(L,R)$ such that

$$\chi\vert _R=\operatorname{id}_R, \qquad L.\chi \subset U$$

holds.

(7): Since $x.\chi$ belongs to $R$ , we know the existence of $r_x$ . The uniqueness of the $r_x$ follows from the assumption that the center of $L$ is trivial.

(8),(9): easy.

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LEMMA 5.61   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

1. The radical $R$ of $L$ is abelian.
2. $p>n^2$ or $p=0$ .

Then $L$ has a Levi subalgebra.

PROOF.. Let $Z$ be the center of $L$ . Applying the previous lemma to $L/Z$ , We see that there exists an Levi subalgebra $L_0/Z$ of $L/Z$ . Now

$$0 \to Z \to L_0 \to L_0/Z \to 0$$

is a short exact sequence of $L_0/Z$ -module, and so it therefore splits. (Theorem 5.53 (Weyl's theorem of irreducibility.)) Thus $L_0$ has a subalgebra $L_1$ which is stable under action of $L_0/Z$ . That means, $L_1$ is a Levi subalgebra of $L_0$ . So $L_1$ is also a Levi subalgebra of $L$ . $\qedsymbol$

THEOREM 5.62 (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 $L$ .

$\qedsymbol$

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