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…)

For the next few weeks, I’m probably going to be doing primarily algebra posts.

Invariant bilinear forms

Let {\mathfrak{g}} be a Lie algebra over the field {k} and {V,W} representations. Recall the following.

1. {v \in V} is invariant under {\mathfrak{g}} if {Xv = 0} for all {X \in \mathfrak{g}}.

2. {\hom_k(V,W)} is a representation of {\mathfrak{g}}: define (Xf)v = X(fv) - f(Xv). This is isomorphic as a {\mathfrak{g}}-module to the tensor product {W \otimes V^{\vee}}, where {V^{\vee}} is regarded as a {\mathfrak{g}}-module. We can think of {W \otimes V^{\vee}} as a {\mathfrak{g}}-module because the enveloping algebra {U\mathfrak{g}} is a Hopf algebra under the homomorphism {U\mathfrak{g} \rightarrow U \mathfrak{g} \otimes U \mathfrak{g}} given by {x \mapsto 1 \otimes x + x \otimes 1} for {x \in \mathfrak{g}}, and extended further.

3. Let {B} be a bilinear form on {V}, i.e. a linear map {B: V \otimes V \rightarrow k}. Then {B} is said to be invariant under {\mathfrak{g}} if for all {v,v' \in V, X \in \mathfrak{g}}

\displaystyle B(Xv, v') + B(v, Xv') = 0;

if we treat {B} as an element of {(V \otimes V)^{\vee}}, this is the same as saying it is invariant in the sense of 1 above.

Ok, all good. Given a representation {V} as above, we have a particular example {B_V} of an invariant and symmetric bilinear form on {\mathfrak{g}} in this post (which in particular shows why some of what I just posted here is redundant; I hadn’t looked back when I started writing it) given by

\displaystyle B_V(x,y) := \mathrm{Tr}( x_V y_V),

where {x_V,y_V} are the corresponding endomorphisms of {V} corresponding to {x,y \in \mathfrak{g}}. An important special case of this is when we are considering the adjoint representation of {\mathfrak{g}} on itself; then this is called the Killing form.

What we shall prove is the following:

Theorem 1 (Cartan)

Let the ground field {k} be of characteristic zero. The Lie algebra {\mathfrak{g}} is solvable if and only if\displaystyle B(\mathfrak{g}, [\mathfrak{g},\mathfrak{g}]) = \{ 0 \},

for {B} the Killing form of the adjoint representation. (more…)

{\mathfrak{sl}_2} is a special Lie algebra, mentioned in my previous post briefly. It is the set of 2-by-2 matrices over {\mathbb{C}} of trace zero, with the Lie bracket defined by:

\displaystyle  [A,B] = AB - BA.

The representation theory of {\mathfrak{sl}_2} is important for several reasons.

  1. It’s elegant.
  2. It introduces important ideas that generalize to the setting of semisimple Lie algebras.
  3. Knowing the theory for {\mathfrak{sl}_2} is useful in the proofs of the general theory, as it is often used as a tool there.

In this way, {\mathfrak{sl}_2} is an ideal example. Thus, I am posting this partially to help myself learn about Lie algebras.