Jordan-Chevalley decomposition of Lie algebra elements

Question

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.)

Answer 1

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!