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
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 2 Let
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 3 An 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 4 Given 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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