Orthogonal complements of semisimple ideals

There is a general fact about semisimple ideals in arbitrary Lie algebras that we prove next; it will be an application of the material on complete reducibility.  We will use it to complete the picture of the abstract Jordan decomposition in a semisimple Lie algebra.

Proposition 1 Let ${\mathfrak{s} \subset \mathfrak{g}}$ be a semisimple ideal in the Lie algebra ${\mathfrak{g}}$. Then there is a unique ideal ${\mathfrak{a} \subset \mathfrak{g}}$ with ${\mathfrak{g} = \mathfrak{s} \oplus \mathfrak{a}}$.

The idea is that we can find a complementary ${\mathfrak{s}}$-module ${A \subset \mathfrak{g}}$ with

$\displaystyle \mathfrak{g} = \mathfrak{s} \oplus A$

as ${\mathfrak{s}}$-modules. Now ${[\mathfrak{s},A] \subset A \cap \mathfrak{s} = \{0\}}$ because ${\mathfrak{s}}$ is an ideal and because ${A}$ is stable under the action of ${\mathfrak{s}}$. The converse is true: ${A}$ is the centralizer of ${\mathfrak{s}}$, and thus an ideal, because anything commuting with ${\mathfrak{s}}$ can have no part in ${\mathfrak{s}}$ in the ${\mathfrak{s} \oplus A}$ decomposition (${\mathfrak{s}}$ being semisimple). Thus, I hereby anoint ${A}$ with the fraktur font and call it ${\mathfrak{a}}$ to recognize its Lie algebraness.

Given a splitting as above, ${\mathfrak{a}}$ would have to be the centralizer of ${\mathfrak{s}}$, so uniqueness is evident. (more…)

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 ${B_D|_{D_i \times D_i} = B_{D_i}}$. (more…)