Wikipedia on the Jordan-Chevalley decomposition (on endomorphisms) says that an analogous decomposition exists in Lie algebras, but no specifics or sources are given. I'm learning about Lie algebras, I don't care about the specifics of how this decomposition is done, but I have this result here that says that for semisimple Lie algebras, if $x=x_s+x_n$ is the Jordan decomposition, that this is consistent with the Jordan decomposition of any representation, i.e. $R(x_s)=R(x)_s$ and $R(x_n)=R(x)_n$. I don't understand what the Jordan decomposition in Lie algebra means. For an operator it's obvious, but Lie algebras are vector spaces, and I don't understand what it means to do a Jordan decomposition on some element $x$ of a Lie algebra. My best guess is that we define some linear map, like the adjoint map $[x,y]$. This seems logical because this is an endomorphism, but I don't know if this is right. (Update: Actually the adjoint map is a derivation, not homomorphism, so it's certainly not this.)
Jordan-Chevalley decomposition of Lie algebra elements
abstract-algebralie-algebrasrepresentation-theory
Related Solutions
It is true as long as the Lie subalgebra is semisimple, and the field has characteristic zero. If not, then there are counterexamples, see this MO question. We also may assume that the field is algebraically closed.
The theorem is as follows. Let $L$ be a linear, semisimple Lie algebra over an algebraically closed field of characteristic zero. Then $L$ contains the semisimple and nilpotent parts in $gl(V )$ of all of its elements. In particular, the abstract and usual Jordan decompositions in $L$ agree.
Proof: Let $x ∈ L$ be given. Let $x = s+n$ be the usual Jordan decomposition in $gl(V)$. The problem is to show $s,n ∈ L$. Since $ad( x)(L) ⊆ L$, we know that both $ad (s)(L) ⊆ L$ and $ad ( n)(L) ⊆ L$. Hence, $s, n ∈ N_{gl(V)}(L) $. In general it is not the case that $N = L$ so we will need to construct a smaller subalgebra containing both $s, n$ which can be shown to equal $L$. To this end we need to use Weyl’s theorem and it’s Lemma that $L ⊆ sl(V )$. Let $W$ be any $L$-submodule of $V$ ,and define $L_W =\lbrace y∈gl(V)|y(W)⊆W, tr(y|W)=0\rbrace$. For example $L_V = sl(V )$. We have $L ⊆ L_W$ , for all W. Define, $L′ = \cap_W (N ∩ L_W ) ⊆ N$. This is a subalgebra which contains $L$, and we will show it is equal to $L$. Before that, because $W$ is $L$-stable, it is in particular $x$-stable and again we have $W$ is stable under both $s, n$. Furthermore, since $0=tr(x)=tr(s+n)$ and $tr(n)=0$ we have $tr(s)=0$ hence $s,n∈L_W$ for all $W$ and thus in $L′$. Hence,it remains to show $L=L′$.From the fact $L′ ⊆N$ we have $[L,L′]⊆L$ so in particular $L′$ is an $L$-module. As $L$ is a submodule in $L′$, Weyl’s Theorem allows us to write $L′ = L ⊕ M$ for some $L$-module $M$. Since $[L, L′] ⊆ L$, we have $L$ acts trivially on $M$. Let $W$ be an irreducible $L$-submodule of $V$. If $y ∈ M$ then, because $[L,y] = 0$, i.e. $y$ commutes with all of $L$, Schur’s Lemma implies that $y$ acts on $W$ as a scalar. On the other hand $y∈M ⊆L′ ⊆L_W$, so we have $tr(y|W)=0$. With the fact $charF = 0$ this implies $y = 0$, hence, $M = 0$, because $V$ is a sum of irreducibles by Weyl’s Theorem, proving $L=L'$.
Best Answer
For semisimple Lie algebras over $\mathbb{C}$, the Jordan decomposition is done in the following way:
Let $\mathfrak{g}$ be a (finite-dimensional) semisimple Lie algebra (over $\mathbb{C}$), and $\operatorname{ad}\colon \mathfrak{g}\to\mathfrak{gl}(\mathfrak{g})$ the adjoint representation. Then the image of $\operatorname{ad}$ is a semisimple subalgebra of $\mathfrak{gl}(\mathfrak{g})$. Note that a semisimple subalgebra $\mathfrak{h}\subseteq \mathfrak{g}$ satisfies the following property: given any $x\in\mathfrak{h}$ (which is a linear operator), the usual Jordan decomposition $x=x_{s}+x_{n}$ satisfies that $x_{s},x_{n}\in\mathfrak{h}$.
Now, let $x\in\mathfrak{g}$. Then we can consider $\operatorname{ad}(x)\in\mathfrak{h}$, and take the Jordan decomposition $\operatorname{ad}(x)=(\operatorname{ad}(x))_{s}+(\operatorname{ad}(x))_{n}$. Since $\operatorname{ad}(\mathfrak{g})$ is semisimple, and $\operatorname{ad}$ is an injective linear map, it follows that there exist unique elements $y,z\in\mathfrak{g}$ such that $\operatorname{ad}(y)=(\operatorname{ad}(x))_{s}$ and $\operatorname{ad}(z)=(\operatorname{ad}(x))_{n}$. Simply define $x_{s}=y$ and $x_{n}=z$. The decomposition $x=x_{s}+x_{n}$ is called the abstract Jordan decomposition of $x$, and one can show that for any representation $\rho\colon\mathfrak{g}\to\mathfrak{gl}(V)$, the decomposition $\rho(x)=\rho(x_{s})+\rho(x_{n})$ is precisely the Jordan decomposition of the linear map $\rho(x)$.
For proofs and more details, you can check Humphreys, Introduction to Lie algebras and Representation Theory.
Hope this helps!