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…)