Theorem of Iwasawa

THEOREM 5.28   Let $L$ be a Lie algebra over a field $k$ of characteristic $p\neq 0$ . Then $L$ has a finite dimensional faithful representation. More precisely, there exists a two-sided ideal $I$ of the universal enveloping algebra $U(L)$ such that $L$ acts faithfully on $U(L)/I$ .

Before proving the above theorem, we first prove the next lemma.

LEMMA 5.29   Under the hypothesis of the theorem, for any $x\in L$ , there exists a monic non constant polynomial $f_x(X) \in k[X]$ such that

$$f_x(x) \in Z(U(L))$$

holds.

PROOF.. Let us put $s=\dim(L)$ . The linear transformation $\operatorname{ad}(x)$ on $L$ is represented by a matrix of size $s$ and has therefore its minimal polynomial $m_x$ : Namely, $m_x$ is a monic polynomial of degree no more than $s$ such that

$$m_x(\operatorname{ad}(x))=0$$

holds. Let us divide $X, X^p, X^{p^2},\dots X^{p^{s+1}}$ by $m_x(X)$ .

$$X^{p^j}=m_x(X) q_j(X)+r_j(X) \qquad(\deg(r_j)<s)\qquad (j=1,2,3,\dots,s+1)$$

Then $s+1$ polynomials $\{r_j(X)\}_{j=1}^{s+1}$ of degree $\leq s-1$ should be linearly dependent. That means, there exists a non trivial vector $(c_j)\in k^{s+1}$ such that

$$\sum_j c_j X^{p^j}\in m_x(X) k[X]$$

holds. Then we have

$$\sum_j c_j (\operatorname{ad}(x))^{p^j}=0.$$

Thus we conclude

$$\operatorname{ad}(\sum_j c_j x^{p^j})=0.$$

By dividing $\sum_j c_j X^{p^j}$ by leading coefficient, we obtain the required polynomial $f_x$ .

$\qedsymbol$

PROOF.. of the Theorem Let $\{e_1,e_2,\dots,e_s\}$ be a basis of $L$ . Then by the above lemma we know that there exists a set of monic non constant polynomials $\{f_1,f_2,\dots, f_s\}$ such that each $h_j=f_j(e_j)$ belongs to the center of $U(L)$ . Let us put $d_j=\deg(f_j)$ . Then using PWB theorem we may easily see that

$$\left\{ h_1^{c_1} h_2^{c_2} h_3^{c_3} \dots h_s^{c_s} e_1^{l_... ... \in \mathbb{N},\\ & l_j<d_j (\forall j) \end{aligned}\right\}$$

forms a basis of $U(L)$ . Let us now put

$$I=U(L) (h_1,\dots,h_s)$$

Then $A=U(L)/I$ is a finite dimensional vector space with the base

$$\left\{ e_1^{l_1} e_2^{l_2} e_3^{l_3} \dots e_s^{l_s} ; \begi... ... \in \mathbb{N},\\ & l_j<d_j (\forall j) \end{aligned}\right\}$$

The representation $\rho_A$ of $L$ on $A$ is faithful. Indeed, for any $x\in L$ , we have

$$\rho_A(x)=0 \implies x.1 =0$$    in $$A \implies x \in I \implies x=0.$$

$\qedsymbol$

DEFINITION 5.30   Let $L$ be a Lie algebra.

1. A representation $V$ of $L$ is called completely reducible if it is a direct sum of reducible sub representations.
2. $L$ is called completely reducible if every representation of $L$ is completely reducible.

The following remark is (at least) in the Book of Bourbaki.

PROPOSITION 5.31   Let $L$ be a non zero finite dimensional Lie algebra over a field $k$ of characteristic $p\neq 0$ . Then $L$ can never be completely reducible.

PROOF.. Let us follow the proof of the theorem of Iwasawa. By taking $f_1^2$ instead of $f_1$ in the proof, we obtain a representation $A_1=U(L)/I$ with a non trivial central nilpotent $z=f_1(e_1)$ . Then we see that $z A_1$ cannot have a direct complementary $L$ -module $X$ . For if it existed, then $X$ should necessarily a left ideal of $A_1$ . On the other hand, by decomposing $1\in A_1$ we obtain

$$1=x + z a \qquad(\exists x\in X \exists a\in A_1).$$

Then $x=1-z a$ has an inverse $(1+z a)$ . This implies that

$$X\supset A_1 x\supset A_1.$$

which is a contradiction. $\qedsymbol$