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.

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 {D_i \subset D} be the algebras of inner derivations (resp. derivations) of a semisimple Lie algebra {\mathfrak{g}}. Then {D_i \simeq \mathfrak{g}} under the {\mathrm{ad}}-homomorphism, so {D} becomes a {\mathfrak{g}}-module. We have then a splitting

\displaystyle D = D_i \oplus \mathfrak{t}

where {\mathfrak{t}} consists of an ideal of derivations that commute with {D_i}. But any derivation that commutes with all of {D_i} is easily checked (as before) to be zero. So {D = D_i}. 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 {x} is a semisimple (resp. nilpotent) element in a semisimple Lie algebra {\mathfrak{g}} and {\phi: \mathfrak{g} \rightarrow \mathfrak{g}'} is a homomorphism of Lie algebras, then {\phi(x)} is semisimple (resp. nilpotent). However, this is not obvious. First:

Proposition 2 Let {\mathfrak{g} \subset gl(V)} be a semisimple subalgebra. Then the abstract and normal Jordan decompositions coincide.

(I follow Humphreys.)

It is sufficient to show that if {x \in \mathfrak{g}}, then the normal semisimple and nilpotent endomorphisms {s, n \in gl(V)} actually lie in {\mathfrak{g}}. For then {\mathrm{ad} s, \mathrm{ad} n} are also semisimple (resp. nilpotent), and the abstract Jordan decomposition is unique.

We will express {\mathfrak{g}} as the intersection of a large amount of subalgebras of {gl(V)} and show that {s,n} belong to each one. First, there is the normalizer {\mathfrak{n} := \{ Y \in gl(V): \mathrm{ad} Y(\mathfrak{g}) \subset \mathfrak{g} \}}. Since {\mathrm{ad} x(\mathfrak{g}) \subset \mathfrak{g}}, we have {\mathrm{ad} s(\mathfrak{g}) \subset \mathfrak{g}} and {\mathrm{ad} n (\mathfrak{g}) \subset \mathfrak{g}} by the lemmas proved here on the normal Jordan decomposition and tensor products. The normalizer will generally be larger than {\mathfrak{g}} though, so we need more subalgebras.

Define for {W \subset V} a subrepresentation of {\mathfrak{g}},

\displaystyle \mathfrak{z}_W := \{ Y \in gl(V): Y(W) \subset W, \ \mathrm{Tr}(Y_W) = 0 \} .

Since {\mathfrak{g} = [\mathfrak{g},\mathfrak{g}]}, the last condition on the trace (along with the first one, of course) is satisfied by {\mathfrak{g}}. It is also satisfied by {n} because {n} is nilpotent, and hence by {s}, and because {n} is a polynomial in {x}, the first condition is also satisfied.

If we thus prove that

\displaystyle \mathfrak{g} := \mathfrak{n} \cap \bigcap_W \mathfrak{z}_W = \mathfrak{g}

we will have proved the proposition. Indeed, {\mathfrak{g}} is an ideal in {\mathfrak{g'}}, so by the first proposition today we can write as Lie algebras

\displaystyle \mathfrak{g}' = \mathfrak{g} \oplus \mathfrak{t}

for {\mathfrak{t}} an ideal.

As a result, any element {t} of {\mathfrak{t}} commutes with {\mathfrak{g}} and is thus a {\mathfrak{g}}-homomorphism on {V}! So {t} induces scalar multiplication on any simple {\mathfrak{g}}-submodule of {V}, which scalar must be zero by the assumption about the trace. In particular, {t =0}, which proves the expression for {\mathfrak{g}} and the proposition.

We have two important corollaries:

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


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

Corollary 4 Given a homomorphism {\phi: \mathfrak{g} \rightarrow \mathfrak{g}'} of semisimple Lie algebras, {\phi} preserves the abstract Jordan decomposition.


Let {x \in \mathfrak{g}} be semisimple. We will show {\phi(x)} is semisimple. The same will apply to nilpotence, which will prove the corollary.

Given a representation of {\mathfrak{g}'}, we pull back via {\phi} to get a representation of {\mathfrak{g}}. Under this second representation {x} must act semisimply; hence under the first one, {\phi(x)} must act semisimply. This proves the corollary.