I learned the material in this post from the book by Humphreys on Lie algebras and representation theory.

Recall that if ${A}$ is any algebra (not necessarily associative), then the derivations of ${A}$ form a Lie algebra ${Der(A)}$, and that if ${A}$ is actually a Lie algebra, then there is a homomorphism ${\mathrm{ad}: A \rightarrow Der(A)}$. In this case, the image of ${\mathrm{ad}}$ is said to consist of inner derivations.

Theorem 1 Any derivation of a semisimple Lie algebra ${\mathfrak{g}}$ is inner.

To see this, consider ${\mathrm{ad}: \mathfrak{g} \rightarrow D :=Der(\mathfrak{g})}$; by semisimplicity this is an injection. Let the image be ${D_i}$, the inner derivations. Next, I claim that ${[D, D_i] \subset D_i}$. Indeed, if ${\delta \in D}$ and ${\mathrm{ad} x \in D_i}$, we have

$\displaystyle [\delta, \mathrm{ad} x] y = \delta( [x,y]) - [x, \delta(y)] = [\delta(x),y] = (\mathrm{ad}(\delta(x)))y.$

In other words, ${[\delta, \mathrm{ad} x] = \mathrm{ad}(\delta(x))}$. This proves the claim.

Consider the Killing form ${B_D}$ on ${D}$ and the Killing form ${B_{D_i}}$ on ${D_i}$. The above claim and the definition as a trace shows that ${B_D|_{D_i \times D_i} = B_{D_i}}$.

Now, if ${D_i^{\perp}}$ is the orthogonal complement to ${D_i}$ under the form ${B_D}$, we must have

$\displaystyle D_i + D_i^{\perp} = D$

but also, because of nondegeneracy of ${B_{D_i}}$,

$\displaystyle D_i \cap D_i^{\perp} = \{ 0 \}.$

I claim now ${D_i^{\perp} = \{0\}}$. If ${\delta \in D_i^{\perp}}$, then for any ${x \in \mathfrak{g}}$,

$\displaystyle [\delta, \mathrm{ad}(x)] = \mathrm{ad}(\delta x) \in D_i \cap D_i^{\perp} = \{ 0 \}$

because ${D_i^{\perp}}$ is an ideal (by invariance of the Killing form again). By semisimplicity, ${\delta x = 0}$, and since ${x}$ was arbitrary, we find ${\delta }$ is the zero derivation. So ${D = D_i}$.

Abstract Jordan decomposition

Now assume our fields are algebraically closed.

Proposition 2 Let ${A}$ be a finite-dimensional algebra. Let ${\delta: A \rightarrow A}$ be a derivation, and regard ${\delta}$ as an element of ${End(A)}$ to take its Jordan decomposition ${\delta = S + N}$. Then ${S,N}$ are derivations of ${A}$ as well.

It is clearly enough to prove ${S}$ is a derivation. There may be confusion caused by my using ${\delta}$ to refer to a specific

The idea is to write ${A}$ as a direct sum

$\displaystyle A = \bigoplus_{\lambda} A_{\lambda}$

where the ${A_\lambda}$ are ${\delta}$-invariant subspaces with ${(\delta-\lambda)}$ acting nilpotently on them. Then ${S}$ acts on ${A_{\lambda}}$ by ${\lambda}$.

It is enough to check that for any ${a \in A_{\lambda}, b \in A_{\mu}}$,

$\displaystyle a b \in A_{\lambda + \mu}.$

For then

$\displaystyle S(ab) = (\lambda + \mu)(ab) = (Sa) b + a(Sb).$

Now indeed, we have a binomial-like formula

$\displaystyle (\delta - \lambda - \mu)^n(ab) = \sum_{i=0}^n \binom{n}{i} (\delta - \lambda)^i a \cdot (\delta - \mu)^{n-i} b$

that can be checked by induction on ${n}$. It shows that ${ab}$ is annihilated by some high power of ${\delta - \lambda - \mu}$.

Theorem 3 (Abstract Jordan decomposition) Let ${\mathfrak{g}}$ be a semisimple Lie algebra. Then we can write any ${x \in \mathfrak{g}}$ uniquely as ${x = s + n}$ where ${\mathrm{ad} s}$ is semisimple, ${\mathrm{ad} n}$ is nilpotent, and ${s,n}$ commute with each other and with every element of ${\mathfrak{g}}$ that commutes with ${x}$.

We imbed ${\mathfrak{g}}$ as a subalgebra ${D_i}$ of ${gl(\mathfrak{g})}$ via the ${\mathrm{ad}}$ mapping. Then ${\mathrm{ad} x \in gl(\mathfrak{g})}$ splits into semisimple and nilpotent parts, ${\mathrm{ad} x = S + N}$. Then ${S,N}$ are derivations of ${\mathfrak{g}}$ by the proposition, and inner ones by Theorem 1, so we get

$\displaystyle \mathrm{ad} x = \mathrm{ad} s + \mathrm{ad} n$

and semisimplicity implies then ${x = s + n}$. Since

$\displaystyle \mathrm{ad} [s,n] = [\mathrm{ad} s, \mathrm{ad} n] = 0,$

we find ${[s,n]=0}$. The commuting properties are thus seen to follow from the corresponding ones of the normal Jordan decomposition. Uniqueness follows by the uniqueness of the Jordan decomposition in ${End(\mathfrak{g})}$.

Note that if we have decompositions ${x = s+n, y = s' + n'}$, and ${x,y}$ commute, then

$\displaystyle x+ y = (s+s') + (n + n')$

is the Jordan decomposition for ${x+y}$.

This by itself is not all that interesting. But it turns out to be the case that semisimple elements (i.e., those whose nilpotent part is zero) in a Lie algebra act by semisimple endomorphisms in any representation. We need to talk about complete reducibility first though.