**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 1Let be a semisimple ideal in the Lie algebra . Then there is a unique ideal with .

The idea is that we can find a complementary -module with

as -modules. Now because is an ideal and because is stable under the action of . The converse is true: is the centralizer of , and thus an ideal, because anything commuting with can have no part in in the decomposition ( being semisimple). Thus, I hereby anoint with the fraktur font and call it to recognize its Lie algebraness.

Given a splitting as above, would have to be the centralizer of , so uniqueness is evident.

**Derivations of semisimple Lie algebras revisited **

Following Serre, we can reprove the innerness of all derivations of a semisimple Lie algebra, just for fun.

Let be the algebras of inner derivations (resp. derivations) of a semisimple Lie algebra . Then under the -homomorphism, so becomes a -module. We have then a splitting

where consists of an ideal of derivations that commute with . But any derivation that commutes with all of is easily checked (as before) to be zero. So . We have thus gotten a quicker version of the previous proof. Before, we needed bilinear forms to get the splitting. Now, we can just appeal to the general splitting criterion.

**More remarks on semisimplicity and nilpotence **

The first thing that should be true is that if is a semisimple (resp. nilpotent) element in a semisimple Lie algebra and is a homomorphism of Lie algebras, then is semisimple (resp. nilpotent). However, this is not obvious. First:

Proposition 2Let be a semisimple subalgebra. Then the abstract and normal Jordan decompositions coincide.

(I follow Humphreys.)

It is sufficient to show that if , then the normal semisimple and nilpotent endomorphisms actually lie in . For then are also semisimple (resp. nilpotent), and the abstract Jordan decomposition is unique.

We will express as the intersection of a large amount of subalgebras of and show that belong to each one. First, there is the normalizer . Since , we have and by the lemmas proved here on the normal Jordan decomposition and tensor products. The normalizer will generally be larger than though, so we need more subalgebras.

Define for a subrepresentation of ,

Since , the last condition on the trace (along with the first one, of course) is satisfied by . It is also satisfied by because is nilpotent, and hence by , and because is a polynomial in , the first condition is also satisfied.

If we thus prove that

we will have proved the proposition. Indeed, is an ideal in , so by the first proposition today we can write as Lie algebras

for an ideal.

As a result, any element of commutes with and is thus a -homomorphism on ! So induces scalar multiplication on any simple -submodule of , which scalar must be zero by the assumption about the trace. In particular, , which proves the expression for and the proposition.

We have two important corollaries:

Corollary 3An element in a semisimple Lie algebra is semisimple (resp. nilpotent) if and only if it acts semisimply (resp. nilpotently) on every representation of .

For one direction, use the adjoint representation. For the other, suppose given a representation , i.e. a homomorphism ; let the image be . An element semisimple in maps to one semisimple in since we are just quotienting by some of the simple factors. An element semisimple in is one that is a semisimple endomorphism of , so we get the corollary (same for nilpotence).

Corollary 4Given a homomorphism of semisimple Lie algebras, preserves the abstract Jordan decomposition.

Let be semisimple. We will show is semisimple. The same will apply to nilpotence, which will prove the corollary.

Given a representation of , we pull back via to get a representation of . Under this second representation must act semisimply; hence under the first one, must act semisimply. This proves the corollary.

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