Abstract Jordan Chevalley decomposition

PROPOSITION 5.63   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

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

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

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

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,

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

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

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

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

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

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

$\qedsymbol$

PROPOSITION 5.64   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

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

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

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

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

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

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 5.65   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 $x$

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

with respect to $V_1$ and that

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

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 $x\in L$ ,

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

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

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

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

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

holds. By comparing entries, we obtain

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

Since $L$ has trivial center, we have

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

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

DEFINITION 5.66   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

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

such that

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

is the Jordan Chevalley decomposition.

PROPOSITION 5.67   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

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

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. )

$\qedsymbol$