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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

$\displaystyle 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

$\displaystyle 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)$ .

$\displaystyle 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

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

holds. Then we have

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

Thus we conclude

$\displaystyle \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

\begin{equation*}
\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\}
\end{equation*}

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

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

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

\begin{equation*}
\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\}
\end{equation*}

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

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

$\displaystyle 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

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

which is a contradiction. $ \qedsymbol$


next up previous
Next: Cartan's criterion for solvability(Ccs) Up: generalities in finite dimensional Previous: functoriality of Killing forms
2007-12-19