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Abstract Jordan Chevalley decomposition

PROPOSITION 5.62   Let $ n$ be a positive integer. Let $ L$ be a $ n$ -dimensional semisimple Lie algebra over a field $ k$ of characteristic $ p\in \operatorname{Ccs}(n^4)$ . Then any derivation $ D\in \operatorname{Der}_k(L)$ of $ L$ is inner. That is, there exists an element $ x=x_D$ such that

$\displaystyle D(y)=[x_D,y]=\operatorname{ad}({x_D})(y).
$

PROOF.. $ \operatorname{Der}_k(L)$ is itself a Lie algebra. Sending each element $ x$ of $ L$ to its ``inner derivation'' $ \operatorname{ad}(x)$ , we obtain a Lie algebra homomorphism

$\displaystyle \operatorname{ad}:L\to \operatorname{Der}_k(L)
$

We note that $ \dim(\operatorname{Der}_k(L)) \leq \dim(L)^2$ , and that $ \operatorname{ad}$ may be viewed as a homomorphism of $ L$ -modules. ($ L$ acts on $ \operatorname{Der}_k(L)$ via $ \operatorname{ad}$ . Namely,

$\displaystyle x.D=\operatorname{ad}(x).D=[\operatorname{ad}(x),D]=[x,D\bullet]-D([x,\bullet])=-\operatorname{ad}(D.x)
$

holds for any $ x\in L$ and for any $ D\in \operatorname{Der}_k(L)$ .) By the Weyl's theorem on complete reducibility, we see that there exists a direct sum decomposition

$\displaystyle \operatorname{Der}_k(L)=\operatorname{ad}(L) \oplus X
$

of $ L$ -modules. Then for any $ D\in X$ and for any $ x\in L$ , we see that

$\displaystyle x.D(=-\operatorname{ad}(D.x) )\in X \cap \operatorname{ad}(L)=0.
$

So $ D=0$ . That means, $ X=0$ .

$ \qedsymbol$

PROPOSITION 5.63   Let $ n$ be a positive number Let $ k$ be a separably closed field of characteristic $ p\in \operatorname{Ccs}(n^4)$ . We assume further that $ n$ is invertible in $ k$ . (This assumption is provided just in case: it probably is not necessary because the assumption $ p\in \operatorname{Ccs}(n^4)$ is presumably much stronger.) Let $ L\subset \mathfrak{gl}_n(k)$ be a linear semisimple Lie algebra. We assume that the representation $ L{}^\curvearrowright k^n$ is irreducible. Then for any element $ x\in L$ , its semisimple part $ x_s$ and its nilpotent part $ x_n$ in $ \mathfrak{gl}_n(k)$ lies in $ L$ .

PROOF.. We may assume $ k$ is algebraically closed. Let $ x\in L$ It is enough to prove $ x_n\in L$ . There exists a polynomial $ f\in k[X]$ such that $ x_n=f(x)$ . Thus we see

$\displaystyle \operatorname{ad}x_n (L) \subset L.
$

Thus $ \operatorname{ad}x_n$ is a derivation of $ L$ . By the preceding lemma we see that there exists an element $ y\in L$ such that

$\displaystyle \operatorname{ad}x_n=\operatorname{ad}y
$

By Schur's lemma, we see that there exists a constant $ c\in k$ such that

$\displaystyle x_n=y+c\cdot 1_n.
$

Let us compute traces of both hand sides. Since $ L=[L,L]$ ($ L$ has no non-trivial ideals.), we have $ \operatorname{tr}(y)=0$ . Since $ x_n$ is nilpotent, we have $ \operatorname{tr}(x_n)=0$ . Thus we conclude $ c=0$ (as we assumed $ n$ is invertible in $ k$ .) $ \qedsymbol$

PROPOSITION 5.64   Let $ n$ be a positive integer. Let $ L$ be a semisimple Lie algebra over a separably closed field $ k$ of characteristic $ p\in \operatorname{Ccs}(n^4)$ . Let $ V_1=(V_1,\pi_1),V_2=(V_2,\pi_2)$ be faithful irreducible representations of $ L$ with dimensions less than or equal to $ n$ . Then for any $ x\in L$ , the Jordan Chevalley decomposition of $ x$

$\displaystyle x=x_s^{(1)}+ x_n^{(1)}
$

with respect to $ V_1$ and that

$\displaystyle x=x_s^{(2)}+ x_n^{(2)}
$

with respect to $ V_2$ coincides.

PROOF.. We consider a faithful representation $ (V,\pi)=(V_1\oplus V_2,\pi_1\oplus \pi_2)$ . For any $ x\in L$ ,

$\displaystyle \pi(x)=
\begin{pmatrix}
\pi_1(x) & 0 \\
0 & \pi_2(x)
\end{pmatri...
...ix}+
\begin{pmatrix}
\pi_1(x_n^{(1)}) & 0\\
0& \pi_2(x_n^{(2)} )
\end{pmatrix}$

satisfies the requirement for the Jordan Chevalley decomposition so by the uniqueness we see

$\displaystyle \pi(x)_s=
\begin{pmatrix}
\pi_1(x_s^{(1)}) & 0\\
0 & \pi_2(x_s^{...
...n=
\begin{pmatrix}
\pi_1(x_n^{(1)}) & 0\\
0& \pi_2(x_n^{(2)} )
\end{pmatrix}.
$

Now we argue in a same way as in the proof of the previous proposition and see that there exists a unique element $ y\in L$ such that

$\displaystyle \operatorname{ad}(x_n)= \operatorname{ad}(y)
$

holds. By comparing entries, we obtain

$\displaystyle \operatorname{ad}(\pi_1(y))=\operatorname{ad}(\pi_1(x_n^{(1)})) ,\quad
\operatorname{ad}(\pi_2(y))=\operatorname{ad}(\pi_2(x_n^{(2)})).
$

Since $ L$ has trivial center, we have

$\displaystyle \pi_1(y)=\pi_1(x_n^{(1)}) ,\quad
\pi_2(y)=\pi_2(x_n^{(2)}).
$

Thus $ y=x_n^{(1)}=x_n^{(2)}$ . $ \qedsymbol$

DEFINITION 5.65   Let $ n$ be a positive integer. Let $ L$ be an $ n$ -dimensional semisimple Lie algebra over a separably closed field $ k$ of characteristic $ p\in \operatorname{Ccs}(n^4)$ . Then the abstract Jordan Chevalley decomposition of $ x$ is an decomposition

$\displaystyle x=x_s +x_n \qquad (x_s,x_n \in L)
$

such that

$\displaystyle \operatorname{ad}(x)=\operatorname{ad}(x_s) +\operatorname{ad}(x_n)
$

is the Jordan Chevalley decomposition.

PROPOSITION 5.66   Let $ n$ be a positive integer. Let $ L$ be an $ n$ -dimensional Lie algebra over a separably closed field $ k$ of characteristic $ p\in \operatorname{Ccs}(n^4)$ Then the abstract Jordan Chevalley decomposition of $ x$ exists. If furthermore there is given a $ m$ -dimensional representation $ (V,\pi)$ of $ L$ and $ p\in \operatorname{Ccs}(m^4)$ , then

$\displaystyle \pi(x)=\pi(x_s)+\pi(x_n)
$

gives the Jordan Chevalley decomposition of $ x$ .

PROOF.. Easy exercise. (Be sure to use Weyl's theorem of complete reducibility. By taking quotient by a certain ideals (kernels of representations) one may reduce the proposition to a case where $ L$ is semisimple and $ \pi$ is faithful and irreducible. )

$ \qedsymbol$


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Next: Bibliography Up: generalities in finite dimensional Previous: Levi decomposition
2007-12-19